(i) Elaborate view of the Galaxies, Planets, stars and other heavenly bodies.
(ii) Camera can be attached for taking photograph for the celestial objects.
(iii) Telescope can be viewed even with the low intensity of light.

(i) Mie scattering takes place when the diameter of the scatterer is similar to or larger than the wavelength of the incident light. It is also an elastic scattering. The amount of scattering is independent of wave length.
(ii) Mie scattering is caused by pollen, dust, smoke, water droplets, and other particles in the lower portion of the atmosphere.

(i) The focal length of eye lens is reduced or the distance between eye lens and retina increases.
(ii) Hence the far point will not be infinity for such eyes and the far point has come closer, with this defect, nearby object can be seen clearly but distant objects cannot be seen clearly.

Snell's law state that, the ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media.
\(\frac{\sin i}{\sin r}=\frac{\mu_2}{\mu_1}\)

(i) Simple Microscope: Simple microscope is a convex lens of short focal length. It is held near the eye to get enlarged image of small objects.
(ii) Let an object (AB) of height 'h' is placed at a point within the principal focus (u < f) of the convex lens and the observer's eye placed just behind the lens.
(iii) As per this position the convex lens CJ produces an erect, virtual and enlarged image (A'B'). The image formed is in the same side of the object and the distance equal to the least distance of distinct vision (D) (D = 25 cm for normal human eye).

(iv) Magnification power (M): Ratio of the height of the image produced by the microscope h₂) to the original height of the object (h₁) is called magnification.
M=\(\frac { A'B' }{ AB } \)
Magnification M =\(\frac { height\ of\ the\ image }{ height\ of\ the\ object } \)
=\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } \)
Magnification can also be defined as the ratio of distance of image (v) to the distance of object (u)
Magnification M =\(\frac { Distance\ of\ the\ image }{ Distance\ of\ the\ object } \)
=\(\frac { v }{ u } \)
Since, the image is virtual and erect; the Magnification (M) is taken as positive In simple Microscope,
when the image is formed at the near point
M=1+\(\frac { D }{ f } \)
when the image is formed at infinity
M=\(\frac { D }{ f } \).

(i) The transparent layer cornea bends the light rays through pupil located at the centre part of the Iris.
(ii) The adjusted light passes through the eye lens. Eye lens is convex in nature. So, the light rays from the objects are converged and a real and inverted image is formed on retina.
(iii) Then, retina passes the received real and inverted image to the brain through optical nerves. Finally, the brain senses it as erect image.

Given:
\(\mathrm{f}=-15 \mathrm{~cm}, \mathrm{u}=-10 \mathrm{~cm}, \mathrm{v}=?, \mathrm{~h}=3 \mathrm{~cm} \)
\(\frac{1}{f} =\frac{1}{v}-\frac{1}{u}  \Rightarrow \frac{1}{v}-\frac{1}{u} =\frac{1}{f} \)
\(\frac{1}{v} =\frac{1}{f}+\frac{1}{u}  \)

\(\frac{1}{v}=\frac{-5}{30}=-\frac{1}{6} \)
\(v=-6 \mathrm{~cm} \)
Size of the image:
Magnification \(=\frac{+v}{u} \quad\left(m=\frac{h^{\prime}}{h}\right)  \)
\(\frac{h^{\prime}}{h} =\frac{-v}{u}  \)
\(\frac{h^{\prime}}{3} =\frac{-6}{-10}=0.6  \)
\(h^{\prime} =0.6 \times 3=1.8 \mathrm{~cm}\)
Size of the image =1.8 cm

a) Yes, he got the image. But the image is not clear.
b) There is no change in the focal length. Because, no change in radius of curvature.

Concave lens should be used to correct myopic eye.
Object distance u = Infinity
Far position of the defective eye v = -90 cm
To find: Power of the lens, P = ?
Nature of the lens =?
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\frac { 1 }{ f } =\frac { 1 }{ -90 } +\frac { 1 }{ \infty } \)
\(\frac { 1 }{ f } =\frac { 1 }{ -90 } \)
∴ f=-90 cm (or) 0.9 m
Power of the lens, P \(\frac { 1 }{ f } =\frac { 1 }{ 0.9 } \) =1.11 D
Concave lens of 1.11D is used to see distance object.