(a) Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope. (b) An astronomical telescope has an objective lens of focal length $20\ \mathrm{m}$ and an eyepiece of focal length $1\ \mathrm{cm}$. (i) Find the angular magnification of the telescope. (ii) If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is $3.5\times10^{6}\ \mathrm{m}$ and the radius of the lunar orbit is $3.8\times10^{8}\ \mathrm{m}$.

(a) Resolving power $=\dfrac{D}{1.22\,\lambda}$. (b)(i) Angular magnification $=2000$. (ii) Diameter of the image $\approx0.18\ \mathrm{m}$.

CBSE Class 12 · 2020 · Geometrical OpticsLong Answer

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(a) Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope. (b) An astronomical telescope has an objective lens of focal length

20 m

and an eyepiece of focal length

1 cm

. (i) Find the angular magnification of the telescope. (ii) If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is

3.5 \times 1 0^{6} m

and the radius of the lunar orbit is

3.8 \times 1 0^{8} m

Solution

(a) In normal adjustment the objective forms the image of the distant object at its focus, which coincides with the focus of the eyepiece, giving a final image at infinity. Resolving power

= \frac{D}{1.22 λ}

.
(b)(i)

m = \frac{f _{o}}{f _{e}} = \frac{20 m}{1 0 ^{- 2} m} = 2000

.
(ii)

d = \frac{diameter of Moon}{radius of orbit} \times f_{o} = \frac{3.5 \times 1 0 ^{6}}{3.8 \times 1 0 ^{8}} \times 20 = 0.18 m

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CBSE Class 12 · 2020 · Geometrical OpticsLong Answer

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