Wave

Given: $f=24$ cm $=0.24$ m

$P=\frac{1}{f}=\frac{1}{0.24}=4.17\text{ D}$

The power is about +4.17 D (about 4.2 D).

$f=\frac{1}{P}=\frac{1}{4}=0.25\text{ m}=25\text{ cm}$

Given: $f = 20\ \text{cm} = 0.2\ \text{m}$.

$P=\frac{1}{f}=\frac{1}{0.2}=+5\ \text{D}$

The power is +5 D. The power is positive, so the lens is a convex (converging) lens.

The student has myopia (short sight), corrected with a concave lens.

Myopia (no lens)                       Corrected with a concave lens

 parallel rays from       eye lens      retina      concave lens     eye lens     retina
 a distant object                          |
 ----------\                              |         ----\   _
 -----------\__  (  )--.                  |         -----)(--\__  (  )-------->* |
 -----------/      (  )   `--*  (focus     |         ----/        (  )     image on
 ----------/                  in front)    |                                retina

The rays are brought to a focus in front of the retina in myopia. A concave lens spreads the rays slightly (diverges them) before they enter the eye, so the eye lens now focuses them exactly on the retina.

(a) Here $f=2$ cm and $u=3.5$ cm, so the object lies between $F$ ($2$ cm) and $2F$ ($4$ cm).

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{2}-\frac{1}{3.5}=\frac{1.5}{7}\;\Rightarrow\; v=\frac{14}{3}\approx 4.67\text{ cm}$

The image forms on the other side of the lens, $4.67$ cm from it, beyond $2F'$. It is real, inverted and magnified.

Ray diagram: draw the principal axis with the lens at $O$. Mark $F$ and $2F$ at $2$ cm and $4$ cm on the left, and $F'$ and $2F'$ at $2$ cm and $4$ cm on the right. Draw the object $AB$ standing on the axis at $3.5$ cm on the left. Ray 1 leaves the top $A$ parallel to the axis and, after refraction, passes through $F'$. Ray 2 leaves $A$ through the optical centre $O$ and goes straight on. The two refracted rays meet beyond $2F'$ below the axis, at $A'$. Drop a perpendicular from $A'$ to the axis to get the inverted image $A'B'$.

(b) Magnification:

$m=\frac{v}{u}=\frac{4.67}{3.5}=1.33$

The image is $1.33$ times as large as the object.

(c) For $u=1.5$ cm (less than $f$):

$\frac{1}{v}=\frac{1}{2}-\frac{1}{1.5}=\frac{1.5-2}{3}=-\frac{1}{6}\;\Rightarrow\; v=-6\text{ cm}$

The image is virtual, erect and magnified ($m=6/1.5=4$). It forms on the same side of the lens as the object, $6$ cm from the lens.

Given: $f=3$ cm, $u=4.5$ cm. The object lies between $F$ ($3$ cm) and $2F$ ($6$ cm).

(a) Ray diagram: draw the axis with the lens at $O$. Mark $F$, $2F$ at $3$ cm and $6$ cm on the left, and $F'$, $2F'$ at $3$ cm and $6$ cm on the right. Draw the object $AB$ at $4.5$ cm on the left. Ray 1 from $A$ travels parallel to the axis and passes through $F'$ after refraction. Ray 2 from $A$ passes through $O$ without deviation. They meet beyond $2F'$ below the axis at $A'$. Drop a perpendicular to the axis to get the inverted image $A'B'$.

(b) $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{3}-\frac{1}{4.5}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\;\Rightarrow\; v=9\text{ cm}$

The image forms $9$ cm from the lens on the opposite side of the object, beyond $2F'$. It is real, inverted and magnified.

$m=\frac{v}{u}=\frac{9}{4.5}=2$

(c) For $u=1.5$ cm (less than $f$):

$\frac{1}{v}=\frac{1}{3}-\frac{1}{1.5}=\frac{1-2}{3}=-\frac{1}{3}\;\Rightarrow\; v=-3\text{ cm}$

The image is virtual, erect and magnified ($m=3/1.5=2$). It forms on the same side of the lens as the object, $3$ cm from the lens.

(b) $v_r > v_y > v_b > v_v$

In glass, the refractive index is least for red and greatest for violet, so red travels fastest and violet slowest ($n=c/v$).

(d) Real, inverted and magnified

When the object is placed between $2F$ and $F$ of a convex lens, the image forms beyond $2F$ on the other side. It is real, inverted and magnified.

(d) A - Red, C - Violet

Red has the longest wavelength and is deviated least, so it lies at one end. Violet has the shortest wavelength and is deviated most, so it lies at the other end, nearest the base of the prism.

Mirage (refraction followed by total internal reflection of light).

Hot air near the road is less dense than the cooler air above it. Light from the sky bends and is totally internally reflected, so the road seems to have water on it.

(i) Differences

(ii) Given: $P=+4$ D

$f=\frac{1}{P}=\frac{1}{4}\text{ m}=0.25\text{ m}=25\text{ cm}$

The focal length is $25$ cm. The power is positive, so the lens is a convex lens.

(i) Concave lens and convex lens

(ii) Denser medium and rarer medium

The student has hypermetropia, removed by spectacles with a convex lens.

Hypermetropia (no lens)                Corrected with a convex lens

 rays from a near         eye lens      retina      convex lens      eye lens     retina
 object (25 cm)                             |
 ----------\                              |         ----\   _
 -----------\__  (  )----------|--*       |         -----()--\__  (  )-------->* |
 -----------/      (  )         | (focus   |         ----/        (  )     image on
 ----------/                    | behind)  |                                retina

The rays would come to a focus behind the retina in hypermetropia. A convex lens converges the rays slightly before they enter the eye, so the eye lens focuses them exactly on the retina.

(a) The friend has hypermetropia (long sight). One reason: the eyeball is too short (or the eye lens has too long a focal length).

(b) A convex lens. It converges the rays from the book a little before they reach the eye, so the eye lens can focus the image on the retina and the book can be read clearly at 25 cm.

Hypermetropia (no lens)                Corrected with a convex lens

 rays from a near         eye lens      retina      convex lens      eye lens     retina
 object (25 cm)                             |
 ----------\                              |         ----\   _
 -----------\__  (  )----------|--*       |         -----()--\__  (  )-------->* |
 -----------/      (  )         | (focus   |         ----/        (  )     image on
 ----------/                    | behind)  |                                retina

The rays would come to a focus behind the retina in hypermetropia. A convex lens converges the rays slightly before they enter the eye, so the eye lens focuses them exactly on the retina.

Defect: Myopia (short sight). The near book is clear, but the far board is blurred.

Causes:

1. The eyeball is too long.
2. The eye lens is too thick (too much curvature), so its focal length is too short.

Remedy: a concave lens, shown below.

Myopia (no lens)                       Corrected with a concave lens

 parallel rays from       eye lens      retina      concave lens     eye lens     retina
 a distant object                          |
 ----------\                              |         ----\   _
 -----------\__  (  )--.                  |         -----)(--\__  (  )-------->* |
 -----------/      (  )   `--*  (focus     |         ----/        (  )     image on
 ----------/                  in front)    |                                retina

The rays are brought to a focus in front of the retina in myopia. A concave lens spreads the rays slightly (diverges them) before they enter the eye, so the eye lens now focuses them exactly on the retina.

(a) He suffers from myopia (short sight).

(b) In the first diagram the rays from the distant board focus in front of the retina. In the second, a concave lens moves the image onto the retina.

Myopia (no lens)                       Corrected with a concave lens

 parallel rays from       eye lens      retina      concave lens     eye lens     retina
 a distant object                          |
 ----------\                              |         ----\   _
 -----------\__  (  )--.                  |         -----)(--\__  (  )-------->* |
 -----------/      (  )   `--*  (focus     |         ----/        (  )     image on
 ----------/                  in front)    |                                retina

The rays are brought to a focus in front of the retina in myopia. A concave lens spreads the rays slightly (diverges them) before they enter the eye, so the eye lens now focuses them exactly on the retina.

(c) One cause: the eyeball is too long (or the eye lens is too thick).

The image forms behind the retina, so the defect is hypermetropia (long sight).

Causes:

1. The eyeball is too short.
2. The eye lens is too thin (its focal length is too long) because the ciliary muscles are weak.

Remedy: use a convex lens in spectacles.

Hypermetropia (no lens)                Corrected with a convex lens

 rays from a near         eye lens      retina      convex lens      eye lens     retina
 object (25 cm)                             |
 ----------\                              |         ----\   _
 -----------\__  (  )----------|--*       |         -----()--\__  (  )-------->* |
 -----------/      (  )         | (focus   |         ----/        (  )     image on
 ----------/                    | behind)  |                                retina

The rays would come to a focus behind the retina in hypermetropia. A convex lens converges the rays slightly before they enter the eye, so the eye lens focuses them exactly on the retina.

(a) Hypermetropia (long sight). The rays focus beyond the retina at A.

(b) One cause: the eyeball is too short (or the eye lens has too large a focal length).

(c) Ray diagram of the correction with a convex lens (right-hand part):

Hypermetropia (no lens)                Corrected with a convex lens

 rays from a near         eye lens      retina      convex lens      eye lens     retina
 object (25 cm)                             |
 ----------\                              |         ----\   _
 -----------\__  (  )----------|--*       |         -----()--\__  (  )-------->* |
 -----------/      (  )         | (focus   |         ----/        (  )     image on
 ----------/                    | behind)  |                                retina

The rays would come to a focus behind the retina in hypermetropia. A convex lens converges the rays slightly before they enter the eye, so the eye lens focuses them exactly on the retina.

(d) A convex lens should be used.

(a) Myopia (short sight). The image forms in front of the retina.

(b) Correction with a concave lens (right-hand part):

Myopia (no lens)                       Corrected with a concave lens

 parallel rays from       eye lens      retina      concave lens     eye lens     retina
 a distant object                          |
 ----------\                              |         ----\   _
 -----------\__  (  )--.                  |         -----)(--\__  (  )-------->* |
 -----------/      (  )   `--*  (focus     |         ----/        (  )     image on
 ----------/                  in front)    |                                retina

The rays are brought to a focus in front of the retina in myopia. A concave lens spreads the rays slightly (diverges them) before they enter the eye, so the eye lens now focuses them exactly on the retina.

(c) A concave lens would be used.

The rays converge in front of the retina, so the defect is myopia.

(a) Causes:

1. The eyeball is too long.
2. The eye lens is too thick (the focal length of the eye lens is too short).

(b) A concave lens is used. It diverges the rays before they enter the eye, so the focus shifts back and the image forms on the retina.

Eye A is myopic (focus in front of the retina). Eye B is hypermetropic (focus behind the retina).

(i) Student B can read the board from the last bench, because a hypermetropic eye sees distant objects clearly.

(ii) Differences in cause

(iii) Eye B is corrected with a convex lens (right-hand part):

Hypermetropia (no lens)                Corrected with a convex lens

 rays from a near         eye lens      retina      convex lens      eye lens     retina
 object (25 cm)                             |
 ----------\                              |         ----\   _
 -----------\__  (  )----------|--*       |         -----()--\__  (  )-------->* |
 -----------/      (  )         | (focus   |         ----/        (  )     image on
 ----------/                    | behind)  |                                retina

The rays would come to a focus behind the retina in hypermetropia. A convex lens converges the rays slightly before they enter the eye, so the eye lens focuses them exactly on the retina.

(ii) Plano concave

The rays focus in front of the retina, which is myopia. Myopia is corrected with a concave lens, and a plano concave lens is one of the concave types.

A concave lens is used to remedy short sightedness (myopia). It diverges the rays so that the image forms on the retina instead of in front of it.

Myopia (no lens)                       Corrected with a concave lens

 parallel rays from       eye lens      retina      concave lens     eye lens     retina
 a distant object                          |
 ----------\                              |         ----\   _
 -----------\__  (  )--.                  |         -----)(--\__  (  )-------->* |
 -----------/      (  )   `--*  (focus     |         ----/        (  )     image on
 ----------/                  in front)    |                                retina

The rays are brought to a focus in front of the retina in myopia. A concave lens spreads the rays slightly (diverges them) before they enter the eye, so the eye lens now focuses them exactly on the retina.

Hypermetropia (long sightedness). A convex lens, used when the object lies within F and 2F, forms a real and magnified image, and this lens corrects hypermetropia.

Wearing contact lenses, or laser surgery (LASIK).

Given: $P=-8$ D

$f=\frac{1}{P}=\frac{1}{-8}\text{ m}=-0.125\text{ m}=-12.5\text{ cm}$

The focal length is $-12.5$ cm. The lens is a concave lens (negative power). It is used for myopia.

Two causes of myopia:

1. The eyeball is too long.
2. The eye lens is too thick (short focal length), so the image forms in front of the retina.

The person has myopia (short sight), because a lens of negative power is a concave lens.

The 40 cm mark is at $2F$, so $2f=40$ cm and $f=20$ cm $=0.2$ m.

$P=\frac{1}{f}=\frac{1}{0.2}=+5\text{ D}$

$f=\frac{1}{P}=\frac{1}{-3}=-0.33\text{ m}=-33.3\text{ cm}$

The focal length is $-33.3$ cm (a concave lens).

$f=\frac{1}{P}=\frac{1}{0.5}=2\text{ m}=200\text{ cm}$

The focal length is +2 m (a convex lens).

(c) $P=\frac{1}{f}$

Power is the reciprocal of focal length in metres, so $f=1/P$ gives the focal length from power. The other options are not formulas for focal length.

Light from the immersed part of the pencil bends away from the normal as it passes from water to air (refraction). The immersed part appears raised, so the pencil seems bent at the water surface.

(b) Due to velocity of light changes when it enters into media having different densities.

The change in speed causes refraction. This makes the spoon look bent, and the water in the aquarium acts like a magnifier so the fish looks larger.

A rainbow is formed by refraction, dispersion and total internal reflection of sunlight inside raindrops.

  Sun's rays (white)
  ===============>  +--------------+
                     |   RAINDROP   |   1. refraction + dispersion
                     |  (  /  \  )  |      on entering the drop
                     |   \      /   |   2. total internal reflection
                     |    '----'    |      at the back of the drop
                     +--------------+   3. second refraction on leaving
                          |      \
                          v       v
                      Violet     Red  (violet bends most, red least)
                          \      /
                           Observer
              (the Sun is behind the observer)

1. Raindrops disperse and reflect sunlight so that red emerges at about $42^\circ$ and violet at about $40^\circ$ to the line from the sun through the observer. All drops giving this angle lie on a cone around that line, which the observer sees as a circular arc. The horizon cuts off the lower part, so a semicircle is seen from the ground.
2. The angular difference between the red and violet rays is always about $2^\circ$, so the band has the same width everywhere.

Diagram

              Prism
   white light   /\          Red    (top, bends least)
  ------------> /  \  ---->  Orange
  (incident ray)/    \----->  Yellow
               /______\----->  Green      Screen
                       ----->  Blue         |
                       ----->  Indigo       |
                       ----->  Violet (bottom, bends most)

Order from top: Red, Orange, Yellow, Green, Blue, Indigo, Violet (VIBGYOR from the base side).

Reason for dispersion: White light has seven colours with different wavelengths. They travel at different speeds in glass, so each has a different refractive index and a different angle of deviation. Red bends least and violet bends most, so the colours separate.

Natural example: the rainbow.

Diagram

              Prism
   white light   /\          Red    (top, bends least)
  ------------> /  \  ---->  Orange
  (incident ray)/    \----->  Yellow
               /______\----->  Green      Screen
                       ----->  Blue         |
                       ----->  Indigo       |
                       ----->  Violet (bottom, bends most)

Order from top: Red, Orange, Yellow, Green, Blue, Indigo, Violet (VIBGYOR from the base side).

Two causes of dispersion:

1. White light is a mixture of seven colours with different wavelengths.
2. Different colours travel at different speeds in glass, so the refractive index of the prism differs for each colour and each bends by a different amount.

              Prism
   white light   /\          Red    (top, bends least)
  ------------> /  \  ---->  Orange
  (incident ray)/    \----->  Yellow
               /______\----->  Green      Screen
                       ----->  Blue         |
                       ----->  Indigo       |
                       ----->  Violet (bottom, bends most)

Order from top: Red, Orange, Yellow, Green, Blue, Indigo, Violet (VIBGYOR from the base side).

(a) Two reasons:

1. Diamond has a very high refractive index (2.42), so its critical angle (about $24.4^\circ$) is very small. Light entering is totally internally reflected many times before leaving.
2. Its cut faces are arranged so light is repeatedly reflected inside and also dispersed into colours. Glass has a lower refractive index and larger critical angle, so it shines less.

(b) Splitting of white light:

              Prism
   white light   /\          Red    (top, bends least)
  ------------> /  \  ---->  Orange
  (incident ray)/    \----->  Yellow
               /______\----->  Green      Screen
                       ----->  Blue         |
                       ----->  Indigo       |
                       ----->  Violet (bottom, bends most)

Order from top: Red, Orange, Yellow, Green, Blue, Indigo, Violet (VIBGYOR from the base side).

(a) Dispersion of light (white light splitting into its colours on passing through a prism).

(b) Ray A is red (deviated least) and ray B is violet (deviated most).

(c) Ray A (red) has more speed in glass, because red has the lowest refractive index.

(d) Violet has a shorter wavelength and travels slower in glass, so its refractive index is higher and it bends more than red.

In the refraction diagram the ray bends away from the normal, so Medium A is denser and Medium B is rarer. Increase the angle of incidence until it exceeds the critical angle, and the ray reflects back into Medium A.

 Medium A (denser)
   incident      \      |      /      reflected
   ray            \  i  |  r   /      ray (r = i)
                   \    |    /
                    \   |   /
 -------------------- O -------------------  boundary
 Medium B (rarer)         | normal (dashed)

Angle of incidence i is greater than the critical angle C (49 degrees for water-air), so there is no refraction and the ray is totally reflected back into Medium A.

 Air (rarer)
 ----------------------------------  boundary
                 | normal
                 |
 Water (denser)  |
        \        |        /
         \  i>C  |  r=i  /
          \      |      /
           \     |     /
            \    |    /
             incident   reflected
             ray        ray (totally reflected)

Angle of incidence i is greater than the critical angle C (49 degrees for water), so there is no refraction and the ray is totally reflected back into the water.

(a) Condition of total internal reflection:

 Air (rarer)
 ----------------------------------  boundary
                 | normal
                 |
 Water (denser)  |
        \        |        /
         \  i>C  |  r=i  /
          \      |      /
           \     |     /
            \    |    /
             incident   reflected
             ray        ray (totally reflected)

Angle of incidence i is greater than the critical angle C (49 degrees for water), so there is no refraction and the ray is totally reflected back into the water.

Light must travel from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle.

(b) Diamond has a very high refractive index (2.42), so its critical angle is very small (about $24.4^\circ$). Light entering it undergoes total internal reflection many times inside before coming out, so it sparkles more than ordinary glass (critical angle about $42^\circ$).

(c) Applications: (i) optical fibres used in communication and in endoscopy; (ii) totally reflecting prisms in periscopes and binoculars; (iii) mirage and brilliance of diamonds.

        A
        |\
        | \
 ray -->|  \   (hypotenuse AC)
 (normal|   \
 incid.)|  45\
        |____\
        B     C

Path: the ray enters face AB along the normal and does not bend. It meets AC at $45^\circ$. This is more than the critical angle of glass ($42^\circ$), so it is totally internally reflected and turns through $90^\circ$ towards face BC. It meets BC along the normal and goes out undeviated, travelling downwards.

          A
          |\
  Object  | \
   ^  --->|  \  AC
   |      |   \
   |      |____\
  ----    B      C
                 
                 Image (inverted, formed below,
                 comes out through BC)

Rays from the object enter face AB along the normal, meet the hypotenuse AC at $45^\circ$ (more than the critical angle $42^\circ$) and are totally internally reflected through $90^\circ$ towards BC. They leave BC without deviation. The image is inverted, the same size as the object, and turned through $90^\circ$.

        A
        |\
  P ---->|  \
        |   \
  P'--->|    \  AC
        |_____\
        B       C
          |  |
          v  v   rays leave through BC, parallel,
                 after total reflection at AC

Both rays PQ and P'Q' enter AB along the normal and are not bent. At AC the angle of incidence is $45^\circ$, more than the critical angle ($42^\circ$), so each is totally internally reflected through $90^\circ$. They emerge from BC perpendicular to it, parallel to each other.

Principle: total internal reflection of light.

(i) The ray strikes the slanting face at $45^\circ$, greater than the critical angle of glass ($42^\circ$), so it is totally internally reflected and turns through $90^\circ$.

   |\
 ->| \   incident at 45 deg
   |  \
   |___\    reflected ray goes down at 90 deg to incident ray

(ii) In water the critical angle is $49^\circ$. The incident ray is at $49^\circ$, so the refracted ray goes along the water surface, with the angle of refraction $90^\circ$.

 Air   ================ surface ===========>  refracted ray grazes surface
                      /|
                     / |normal
 Water              /49|
                   incident ray

(i) The ray enters face AB along the normal, so it is not deviated and travels horizontally. Angle A is $60^\circ$, so the face AC makes $30^\circ$ with the horizontal ray. The angle of incidence (between the ray and the normal to AC) is therefore

$\theta_1 = 90^\circ-30^\circ = 60^\circ$

This is greater than the critical angle of glass ($42^\circ$), so the ray is totally internally reflected.

(ii) The ray reflects at AC and moves downwards at $60^\circ$ below the horizontal. At Q on base BC the angle of incidence is $30^\circ$ with the normal. This is less than the critical angle, so the ray refracts out into the air, bending away from the normal, below the prism.

1. The light must travel from a denser medium to a rarer medium (for example glass to air or water to air).
2. The angle of incidence in the denser medium must be greater than the critical angle for that pair of media.

Optical fibres use total internal reflection to carry light signals (telephone, internet and endoscopy). Other examples are periscopes and binoculars.

(a) The image is virtual and magnified

When an object is between the focus F and the optical centre of a convex lens, the image is virtual, erect and magnified. It forms on the same side as the object.

(i) Ray diagram (object PQ within F):

          Q'  (virtual, erect, magnified image)
         /|
        / |        Q
       /  |       /|\
 -----/---+-----F-/-+--------|------- principal axis
     P'   |    P   |        O  (lens)
                   
 Ray 1: Q -> parallel to axis -> lens -> passes through F on the right side
 Ray 2: Q -> through O -> goes straight
 Refracted rays diverge; their backward extensions meet at Q' on the same side as the object.

Object PQ lies between F and the lens; image P'Q' is on the same side, farther from the lens and larger.

(ii) The image is virtual, erect and magnified (larger than the object), formed on the same side of the lens as the object.

Object at 2F:

            Q
            |\
   ---------|-\-------------------_______ 
  Q---------+--\----(lens)       ______/
            |    \ 
 --2F-------F-----O-----F-------2F-------
      P                        

Object at 2F: ray 1 (parallel to axis) refracts through F on the far side; ray 2 (through O) passes straight. They meet at 2F on the other side: image P'Q' is real, inverted, same size as the object.

Character of the image: real, inverted and equal in size to the object.

The object is at $2F$:

            Q
            |\
   ---------|-\-------------------_______ 
  Q---------+--\----(lens)       ______/
            |    \ 
 --2F-------F-----O-----F-------2F-------
      P                        

Object at 2F: ray 1 (parallel to axis) refracts through F on the far side; ray 2 (through O) passes straight. They meet at 2F on the other side: image P'Q' is real, inverted, same size as the object.

(a) Object A is between $2F$ (6 cm) and $F$ (3 cm):

 Q            ray 1 (parallel to axis)          
 |  -----------------------\__ 
 |                          (lens)\___ passes through F
 P-------2F--------F--------O--------F--------2F-------P'
                                              \  
                                               Q' (inverted image, beyond 2F)

Ray 1 from Q parallel to the axis goes through F after refraction. Ray 2 through O goes straight. They meet beyond 2F at Q'. The image is real, inverted and magnified.

(b) The image is real and inverted (and magnified, formed beyond $2F$).

(c) At 2.5 cm the object is inside the focus ($2.5<3$ cm). The image changes from real and inverted to virtual and erect, and it becomes larger and farther, on the same side as the object.

Given: $u=3$ cm, $f=2$ cm. The object is between $F$ and $2F$ ($2<3<4$).

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\Rightarrow v=6\text{ cm}$

 Q            ray 1 (parallel to axis)          
 |  -----------------------\__ 
 |                          (lens)\___ passes through F
 P-------2F--------F--------O--------F--------2F-------P'
                                              \  
                                               Q' (inverted image, beyond 2F)

Ray 1 from Q parallel to the axis goes through F after refraction. Ray 2 through O goes straight. They meet beyond 2F at Q'. The image is real, inverted and magnified.

The image forms 6 cm from the lens (beyond $2F=4$ cm). It is real, inverted and magnified ($m=v/u=2$).

Given: $f=2.5$ cm, $u=6$ cm. Since $u>2f=5$ cm, the object is beyond $2F$.

$\frac{1}{v}=\frac{1}{2.5}-\frac{1}{6}\Rightarrow v\approx4.3\text{ cm}$

 Q
 |    ray 1 (parallel to axis)
 |  ------------------------\
 |                           ( )-----\_______ through F
 P----------2F-------F------O----F--P'--2F---
                                      \   
                                       Q'  (between F and 2F)

The image is formed between $F$ and $2F$ on the other side. It is real, inverted and diminished.

Given: $f=4$ cm, $u=8.5$ cm ($u>2f=8$ cm).

$\frac{1}{v}=\frac{1}{4}-\frac{1}{8.5}\Rightarrow v\approx7.6\text{ cm}$

 Q
 |    ray 1 (parallel to axis)
 |  ------------------------\
 |                           ( )-----\_______ through F
 P----------2F-------F------O----F--P'--2F---
                                      \   
                                       Q'  (between F and 2F)

The image is formed between $F$ and $2F$. It is real, inverted and diminished.

Power: $f=4$ cm $=0.04$ m

$P=\frac{1}{f}=\frac{1}{0.04}=+25\text{ D}$

(i) Ray diagram:

 Q   ray 1 (parallel to axis)
 |  --------------------\      diverges as if from F (left)
 |                      ( ) ------->
 P--F---------------------O--------F--------
          Q' (image between P' and lens, same side)

Ray 1 from Q parallel to the axis diverges after the lens as if it came from the focus on the object's side. The ray through O goes straight. Their backward extensions meet at Q'. The image is virtual, erect and diminished, between the object and the lens.

(ii) No. A concave lens forms a diminished image, so it cannot be used to observe the parts of a flower. A convex lens (magnifying glass) is needed.

(iii) $f=10$ cm $=0.1$ m. For a concave lens $f$ is negative, so

$P=\frac{1}{f}=\frac{1}{-0.1}=-10\text{ D}$

(iv) With a lens thicker at the middle (a convex lens), an object within $F$ gives an image that is virtual and erect, but magnified instead of diminished.

(i) Lenses A and B are convex lenses and can be used to correct long sightedness (hypermetropia).

(ii) $f=3$ cm, object between $F$ and $O$ (say $u=2$ cm).

          Q'  (virtual, erect, magnified image)
         /|
        / |        Q
       /  |       /|\
 -----/---+-----F-/-+--------|------- principal axis
     P'   |    P   |        O  (lens)
                   
 Ray 1: Q -> parallel to axis -> lens -> passes through F on the right side
 Ray 2: Q -> through O -> goes straight
 Refracted rays diverge; their backward extensions meet at Q' on the same side as the object.

Object PQ lies between F and the lens; image P'Q' is on the same side, farther from the lens and larger.

The image is virtual, erect and magnified. Check: $\frac{1}{v}=\frac{1}{3}-\frac{1}{2}\Rightarrow v=-6$ cm, so the image is 6 cm from the lens on the object's side.

(iii) Lens C is a concave lens. It forms a virtual, erect and diminished image at every object position, including 2F.

Types of lenses

Lens A (convex) forms a virtual and magnified image when the object is 15 cm away, since $15<f=20$ cm (object within F).

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\ (u=-15)=\frac{1}{20}-\frac{1}{15}\Rightarrow v=-60\text{ cm}$

          Q'  (virtual, erect, magnified image)
         /|
        / |        Q
       /  |       /|\
 -----/---+-----F-/-+--------|------- principal axis
     P'   |    P   |        O  (lens)
                   
 Ray 1: Q -> parallel to axis -> lens -> passes through F on the right side
 Ray 2: Q -> through O -> goes straight
 Refracted rays diverge; their backward extensions meet at Q' on the same side as the object.

Object PQ lies between F and the lens; image P'Q' is on the same side, farther from the lens and larger.

The image is 60 cm from the lens on the same side as the object. It is virtual, erect and magnified.

(b) convex lens

A convex lens is thicker in the middle and thinner at the edges. A concave lens is the opposite.

The thicker the lens (the more curved its surfaces), the shorter its focal length. A thin lens has a long focal length: thickness is inversely related to focal length.

Given: speed of light in air $c=3\times10^{8}$ m/s; speed in water $v=2.25\times10^{8}$ m/s (the printed $10^{5}$ is a misprint for $10^{8}$).

Formula: $n=\dfrac{c}{v}$

$n=\frac{3\times10^{8}}{2.25\times10^{8}}=1.33$

The refractive index of water is 1.33.

Independent variable: the medium (and so the speed of light in it) through which light passes. It is the one we change.

Dependent variable: the refractive index $n$ of that medium. It changes because of the medium chosen.

Relation:

$n=\frac{\text{speed of light in vacuum (air)}}{\text{speed of light in the medium}}=\frac{c}{v}$

The slower the light in a medium, the higher its refractive index and the denser the medium is optically.

(b) $1.24\times10^{8}$ m/s

The refractive index of diamond is 2.42, so $v=c/n=3\times10^{8}/2.42\approx1.24\times10^{8}$ m/s.

(c) R

The distance from the optical centre to the centre of curvature is the radius of curvature, $R$. Also, $R=2f$.

(a) Principal axis

The imaginary straight line passing through the two centres of curvature of a lens is its principal axis.

The new medium is a rarer medium. A ray bends away from the normal (larger angle) when it enters a rarer medium from a denser one.

Medium B. The ray goes from B to A and bends away from the normal, so B is the denser medium and A is rarer. The critical angle exists only in the denser medium, where the angle of incidence can be increased until refraction is at $90^\circ$.

(i) p - angle of refraction and q - critical angle.

In the first diagram p is the refracted ray's angle with the normal. In the second, the refracted ray grazes the surface, so the angle of incidence q is the critical angle.

(d) picture (iv)

The angle of incidence ($52^\circ$) is greater than the critical angle of water ($49^\circ$) and the ray goes from water to air, so there is total internal reflection back into the water.

The ray enters the curved face towards the centre, so it is not bent there. At the flat face the ray is partly reflected, and when the angle of incidence exceeds the critical angle ($42^\circ$ for glass), total internal reflection occurs. Reading from the scale (a typical case for glass):

(The angle of reflection always equals the angle of incidence; read the values from the scale in the figure.)

Difference between rectangular and semicircular slabs

(b) Medium 1 is rarer and medium 2 denser.

The angles to the surface are $30^\circ$ and $52^\circ$, so the angles with the normal are $60^\circ$ in medium 1 and $38^\circ$ in medium 2. The ray makes a smaller angle with the normal in medium 2, so it bends toward the normal, which means medium 2 is denser. Light is therefore faster in medium 1. Option (d) is also true for the same reason, because medium 1 is the rarer medium.

 Air (rarer)             normal
                           |
         incident ray      |
                 \         |
                  \ i      |
 ------------------\-------+------------ boundary
                    \      |
                     \  r  |
 Glass (denser)       \    |
                       v   
                   refracted ray (bends towards the normal)

When light passes from air (rarer) into glass (denser) it bends towards the normal, so $r<i$.

Light travels slower in glass (denser) than in air (rarer). On entering the denser medium, the speed decreases and the ray bends towards the normal.

(iv) $i=e>r$

On entering glass the ray bends toward the normal, so $r<i$. On leaving through the parallel face it bends away by the same amount, so the emergent angle $e=i$. Hence $i=e>r$.

(a) Statement and argument 1 are correct but argument 2 is wrong; Argument 1 is the reason of the statement.

The atmosphere has layers of different density. Sun's rays pass from the rarer upper layers into denser lower ones, so they bend towards the normal, not away from it. This makes the Sun visible about 2 minutes before it actually reaches the horizon. So argument 2 (bending away) is wrong. Options (a) and (c) have the same meaning, and the physics points to (a).

(ii) Wavelength of violet ray is less and frequency is high.

Violet has the shortest wavelength among visible colours, and since $c=f\lambda$, its frequency is the highest.

(i) The condition is called a mirage.

(ii) Main cause: the road surface heats the air just above it, making the lower layers less dense (rarer) than the upper layers. Light from the sky goes from the denser upper layer to the rarer lower layers, bends away from the normal and finally undergoes total internal reflection. The observer sees this as an image of the sky, which looks like water.

(iii) Two conditions:

1. The ground (road or sand) must be very hot, so the air density increases with height.
2. The angle of incidence of light reaching the lower air layers must exceed the critical angle (a clear sky and calm air).

The figure is related to mirage, which is caused by refraction and total internal reflection of light in air layers of different densities.

The figure shows a mirage. It is the optical illusion in which refraction and total internal reflection of light through air layers of different density produce an inverted image of distant objects (such as a tree or the sky), which seem to be on the ground.

Application: the same total internal reflection that forms a mirage is used in optical fibres to carry signals.

The ciliary muscles change the curvature (and so the focal length) of the eye lens, which is how the eye accommodates.

The power of a lens is the reciprocal of its focal length expressed in metres, $P=\dfrac{1}{f}$. Its unit is the dioptre (D).

For a normal eye the near point is 25 cm from the eye, and the far point is at infinity.

The main defects of vision are myopia (short-sightedness), hypermetropia (long-sightedness) and presbyopia. Astigmatism is another defect.

Completing the diagram: Draw the ray from the top of AB parallel to the principal axis; after the lens it passes through the far focus $F$. Draw the ray from the top of AB through the optical centre $O$; it goes straight on. The two refracted rays meet beyond $2F$ on the other side, and the image $A'B'$ is drawn from that point down to the axis.

Nature of the image (any two): it is formed beyond $2F$, it is real, inverted, and magnified.

(a) $f=\frac{1}{P}=\frac{1}{2}\text{ m}=0.5\text{ m}=50\text{ cm}$

The focal length is $50$ cm.

(b) The power is positive, so the lens is a convex lens. A convex lens is used to correct hypermetropia (long sight), so the person has hypermetropia.

(a) The student cannot see distant letters clearly, so the defect is myopia (short sight). A concave lens should be used in the spectacles.

(b) Ray diagram: draw an eye with its lens and retina. In front of it draw a concave (diverging) lens. Draw parallel rays from a distant object falling on the concave lens. After the concave lens the rays diverge, as if coming from the far point of the myopic eye (dotted lines traced back on the object side). The eye lens then converges these rays and they meet exactly on the retina, forming a clear image.

In myopia the eyeball is too long (or the eye lens too thick), so rays from distant objects meet in front of the retina. The concave lens diverges the rays slightly before they enter the eye, so the image shifts back onto the retina and the vision becomes clear.

OR The concave lens diverges the parallel rays coming from a distant object so that they appear to come from the far point of the myopic eye. The eye lens can focus rays from this point on the retina, so the blurred vision is removed.

Given: $h_o=2$ cm, $u=5$ cm, $f=3$ cm.

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\;\Rightarrow\; v=7.5\text{ cm}$

Position and nature: the object lies between $F$ ($3$ cm) and $2F$ ($6$ cm). The image forms on the other side of the lens at $7.5$ cm, beyond $2F$. It is real, inverted and magnified.

Magnification:

$m=\frac{v}{u}=\frac{h_i}{h_o}=\frac{7.5}{5}=1.5$

So the image height is $h_i=1.5\times 2=3$ cm.

Scale ray diagram (scale $1$ cm : $1$ cm): draw the principal axis and a lens at the centre. Mark $F$ and $2F$ at $3$ cm and $6$ cm on both sides. Draw the candle $2$ cm high at $5$ cm on the left. Ray 1 goes parallel to the axis and then through the right focus. Ray 2 passes undeviated through the optical centre. They meet $7.5$ cm to the right and $3$ cm below the axis, which is the tip of the inverted image.

Formula for magnification: $m=\frac{\text{height of image}}{\text{height of object}}=\frac{\text{image distance}}{\text{object distance}}=\frac{v}{u}$

Case 1: $m<1$ (object beyond $2F$)

Ray diagram: draw the axis with the lens at $O$, marking $F$, $2F$ on the left and $F'$, $2F'$ on the right. Place the object $AB$ beyond $2F$. Ray 1 from $A$ runs parallel to the axis and passes through $F'$ after refraction. Ray 2 from $A$ passes undeviated through $O$. They meet between $F'$ and $2F'$ below the axis at $A'$, giving a small, inverted image $A'B'$. The image is real, inverted and diminished.

Case 2: $m>1$ (object between $F$ and $2F$)

Ray diagram: with the same axis and marks, place $AB$ between $F$ and $2F$. Ray 1 from $A$ runs parallel to the axis and passes through $F'$. Ray 2 passes undeviated through $O$. They meet beyond $2F'$ below the axis at $A'$, giving a large, inverted image $A'B'$. The image is real, inverted and magnified.

Common characteristic: in both cases the image is real and inverted (it forms on the opposite side of the lens and can be taken on a screen).

Different characteristic: when $m<1$ the image is diminished (smaller than the object), but when $m>1$ the image is magnified (larger than the object).

(a) Ray diagram: Ray 1 from the top $A$ of the object runs parallel to the principal axis. After the concave lens it diverges as if it came from the focus $F$ on the object's side, so draw it bending away from the axis and extend it backwards with a dotted line to $F$. Ray 2 from $A$ passes straight through the optical centre. The refracted rays do not meet, so extend them backwards; they meet between the lens and $F$ on the object's side at $A'$. Draw the perpendicular from $A'$ to the axis to get the image $A'B'$.

(b) The image is virtual and erect; it is also diminished (smaller than the object). Any two of these.

(c) For a concave lens the power is $-5$ D, so

$f=\frac{1}{P}=\frac{1}{-5}=-0.2\text{ m}=-20\text{ cm}$

The focal length is $20$ cm (the negative sign shows it is a concave lens).

(d) With a convex lens, the image formed from an object between $F$ and $2F$ would be real and inverted instead of virtual and erect.

OR

Natures of images:

For a convex lens of power $+5$ D:

$f=\frac{1}{5}=0.2\text{ m}=20\text{ cm}$

Differences

Given: $P=-6$ D

$f=\frac{1}{P}=\frac{1}{-6}\text{ m}=-0.167\text{ m}=-16.7\text{ cm}$

The focal length is $-16.7$ cm (about $-17$ cm). The negative sign shows the lens is a concave lens, used to correct myopia.

$f=\frac{1}{P}=\frac{1}{1}=1$ m. The focal length is 1 metre (100 cm).

i. Differences between concave and convex lens

ii. Light travels faster in air than in glass. The medium in which light has a higher speed (lower refractive index) is called the rarer medium, so air is rarer. The medium in which light has a lower speed (higher refractive index) is the denser medium, so glass is denser.

A concave lens. The student has myopia (short sight): distant letters are focused in front of the retina. A concave lens diverges the rays so the image forms on the retina.

(i) The defect is hypermetropia (long sight).

(ii) One cause: the eyeball is too short (or the eye lens is too thin, so its focal length is too long). The image of a near object forms behind the retina.

(iii) Use spectacles with a convex lens of suitable power. Contact lenses or surgery are also options.

(iv) Diagram showing the correction:

Hypermetropia (no lens)                Corrected with a convex lens

 rays from a near         eye lens      retina      convex lens      eye lens     retina
 object (25 cm)                             |
 ----------\                              |         ----\   _
 -----------\__  (  )----------|--*       |         -----()--\__  (  )-------->* |
 -----------/      (  )         | (focus   |         ----/        (  )     image on
 ----------/                    | behind)  |                                retina

The rays would come to a focus behind the retina in hypermetropia. A convex lens converges the rays slightly before they enter the eye, so the eye lens focuses them exactly on the retina.

A convex lens. His mother has hypermetropia (long sight), and a convex lens converges the rays so the image of near objects forms on the retina.

Light from the bottom of the pond passes from water (denser) to air (rarer) and bends away from the normal. The rays appear to come from a point higher than the real bottom, so the pond looks shallower than it really is.

Optical fibre is used for fast transmission of information such as telephone, internet and cable TV signals. It is also used in an endoscope to look inside the body.

(i) Ray diagram:

          Q'  (virtual, erect, magnified image)
         /|
        / |        Q
       /  |       /|\
 -----/---+-----F-/-+--------|------- principal axis
     P'   |    P   |        O  (lens)
                   
 Ray 1: Q -> parallel to axis -> lens -> passes through F on the right side
 Ray 2: Q -> through O -> goes straight
 Refracted rays diverge; their backward extensions meet at Q' on the same side as the object.

Object PQ lies between F and the lens; image P'Q' is on the same side, farther from the lens and larger.

(ii) Characteristics of the image: virtual and erect; magnified.

(iii) The device is a magnifying glass (simple microscope). It is used to read small print, or by watchmakers to see small parts of a watch.

The incident ray is the ray that falls on the surface of a medium. The emergent ray is the ray that comes out of the medium after refraction.