The myopic person uses a spectacle of power, P = -1.0 D

Focal length of the given spectacles, f = $\frac{1}{P}$ = $\frac{1}{-1\times 10^{-2}}$

= -100 cm

Therefore, 100cm is the far point of the person. The normal near point of the person is 25cm. When the objects are placed at infinity, virtual images are produced at 100cm. This is possible when the spectacles are used. When ability of accommodation of the eye-lens is used, he will be able to see the objects that are placed between 100cm and 25cm.

During old age, the person uses reading glasses of power, P’ = +2 D

As the age increases, the ability of accommodation decreases. This is known as presbyopia. This is the reason why he is finding it difficult to see the objects placed at 25cm

(i) The two prisms must be place next to each other.  The bases of these two prisms must be such that they are on the opposite sides of the incident white light. The dispersion of white light first takes place when it is incident on the first prism. The dispersed ray then enters the second prism as an incident ray. The dispersion of light from both the prisms emerges as a white light.

(ii) Consider the two prisms from (i). The deviations from this combination of prisms become equal by adjusting the angle between them. When the angle is maintained between these two prisms, the pencil of white light will disperse without much deviation.

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:

$f=\frac{D^{2}-d^{2}}{4D}$

=$\frac{90^{2}-(20^{2})}{4\times90}=\frac{770}{36}=21.39cm$
Therefore, the focal length of the convex lens is 21.39 cm.

Distance between the object and the image, d = 3 m

Maximum focal length of the convex lens =$f_{max}$
For real images, the maximum focal length is given as:

$f_{max}=\frac{d}{4}$

=$\frac{3}{4}=0.75m$
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m

(i) Yes.

Real images can be obtained from plane and convex mirrors too. When the light rays converge at a point behind the plane or convex mirror, the object is said to be virtual. The real image of this object is obtained on the screen which is placed in the front of the mirror. This is where the real image is obtained.
(ii) No

To obtain a virtual image, the light rays must diverge. In the eyes, the convex lens help in converging these divergent rays at the retina. This is an example where the virtual image acts as an object for the lens to produce a real image.

(iii) The diver is in the water while the fisherman is on the land. Air is less dense when compared to the water as a medium. It is mentioned that the diver is viewing the fisherman. This explains that the light rays are travelling from the denser medium to the rarer medium. This means that the refracted rays move away from the normal making fisherman appear taller.

(iv) Yes; Decrease

The reason behind the change in depth of the tank when viewed obliquely is because the light rays bend when they travel from one medium to another. This also means that the apparent depth is less than the near-normal viewing.

(v) Yes

2.42 is the refractive index of the diamond while 1.5 is the refractive index of the ordinary glass. Also, the critical angle of the diamond is less than the glass. The sparkling effect of a diamond is possible because of the large cuts in the angle of incidence. Larger cuts ensures that the light entering the diamond is totally reflected from all the faces

The actual depth of the pin, d = 15 cm
Apparent depth of the pin =d’
Refractive index of glass,$\mu$=1.5
The ratio of actual depth to the apparent depth and the refractive index of the glass are equal.
i.e.

$\mu=\frac{d}{d’}$

Therefore d’=$\frac{d}{\mu}$

= $\frac{15}{1.5}$=10cm
The distance at which the pin appears to be raised =d’-d=15-10=5cm
When the angle of incidence is small, the distance is independent of the location of the slab

(i) The focal length is negative for a concave mirror.
Therefore f < 0
The object distance (u) is negative when the object is placed on the left side of the mirror.
Therefore u < 0
Lens formula for image distance v is written as:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$——————(i)

For a concave mirror, f is negative, i.e., f < 0

For a real object, i.e., which is on the left side of the mirror.

For u between f and 2f implies that 1/u lies between 1/f and 1/2f

$\\\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}(as \, u, \, f \, are \, negative) \\ \\\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0 \\ \\\frac{1}{2f}<\frac{1}{v}<0$

1/v is negative

which implies v is negative and greater than 2f. Hence, image lies beyond 2f and it is real.

(ii) The focal length is positive for a convex mirror.
Therefore f > 0
The object distance (u) will be negative if the object is placed on the left side of the mirror.
Therefore u < 0
Using mirror formula, we can calculate the image distance v:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

Using eqn (ii), we can conclude that:

$\frac{1}{v}<0$

v>0

Thus, the image is obtained on the back side of the mirror, Therefore, it can be concluded that a convex mirror always produces a virtual image irrespective of the object distance.

(iii) The focal length is positive for a convex mirror.
Therefore f > 0
The object distance (u) is negative when the object is placed on the left side of the mirror
Therefore u < 0
For image distance v, we have the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

But we have u<0

Therefore $\frac{1}{v}>\frac{1}{f}$

v

Therefore, the image is formed between the focus and the pole and is diminished.
(iv) The focal length of the concave mirror is negative.

Therefore f < 0

The object distance, u is negative for an object that is placed on the left side of the mirror.
Therefore u < 0
It is placed between the focus (f) and the pole.

Therefore f>u>0

$\frac{1}{f}<\frac{1}{u}<0$

$\frac{1}{f}-\frac{1}{u}<0$
For image distance v, we have the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

Therefore $\frac{1}{v}<0$

v>0
The image obtained is virtual since it is formed on the right side of the mirror.
For u < 0 and v > 0, we can write:

$\frac{1}{u}>\frac{1}{v}$

v>u
Magnification, m=$\frac{v}{u}$> 1
Hence, the formed image is enlarged

Focal length of the objective lens, $f_{o}$=15m =$15 \times 10^{2} cm$
Focal length of the eyepiece, $f_{e}$= 1.0 cm
(i) $\alpha =\frac{f_{o}}{f_{e}}$ is the angular magnification of a telescope.

=$\frac{15\times10^{2}}{1}$=1500
Hence, the angular magnification of the given refracting telescope is 1500.
(ii) Diameter of the moon, d = $3.48 \times 10^{6} m$
Radius of the lunar orbit, $r_{o}$= $3.8 \times 10^{8} m$
Consider d’ to be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by
the image.

$\frac{d}{r_{o}}=\frac{d’}{f_{o}}$

$\frac{3.48\times10^{6}}{3.8\times10^{8}}=\frac{d’}{15}$

Therefore d’=$\frac{3.48}{3.8}\times10^{-2}\times15$

$13.74\times10^{-2}m=13.74cm$

Therefore 13.74cm is the diameter of the image of the moon formed by the objective lens.

Focal length of the objective lens, $f_{o}$= 144 cm
Focal length of the eyepiece, $f_{e}$= 6.0 cm
The magnifying power of the telescope is given as:
m = $\frac{f_{o}}{f_{e}}$

=$\frac{144}{6}$=24

The separation between the objective lens and the eyepiece is calculated as: $f_{o}+f_{e}$=144+6=150cm

Therefore, 24 is the magnifying power of the telescope and the separation between the objective lens and the eyepiece is 150cm.

Focal length of the objective lens,$f_{o}$= 8 mm = 0.8 cm

Focal length of the eyepiece, $f_{e}$= 2.5 cm

Object distance for the objective lens, $u_{o}$ = -9.0 mm = -0.9 cm

Least distance of distant vision, d=25 cm

Image distance for the eyepiece, $v_{e}$ = -d=-25 cm

Object distance for the eyepiece =$u_{e}$.

Using the lens formula, we can obtain the value of $u_{e}$ as:

$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$

$\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}$

$\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}$

$u_{e}=-\frac{25}{11}=-2.27cm$

With the help of lens formula, the value of the image distance for the objective (v) lens is obtained:

$\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}$

$\frac{1}{v_{o}}=\frac{1}{f_{o}}+\frac{1}{u_{o}}$

$=\frac{1}{0.8}-\frac{1}{0.9}$

$=\frac{1}{7.2}$

vo = 7.2 cm

The separation between two lenses are determined as follows:

= v0 + |ue|

= 7.2 + 2.27

= 9.47 cm

The magnifying power of the microscope is calculated as follows:

Magnifying power, M = Mo × Me

$\\M = \frac{v_{o}}{u_{o}}(1+\frac{D}{f_{e}}) \\ \\M = \frac{7.2}{0.9}(1+\frac{25}{25})$

= 8 × 11

= 88