CBSE Class 12 Physics 2019 Delhi Set 1 Paper

Question : 19 of 27

Marks: +1, -0

Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refraction telescope at an observatory has an objective lens of focal length

15\,\mathrm{m}

and an eyepiece of focal length

1.0\,\mathrm{cm}

. If this telescope is used to view the moon, find the diameter of the image of the moon formed by the objective lens. The diameter of the moon is

3.48 \times 10^6\,\mathrm{m}

, and the radius of lunar orbit is

3.8 \times 10^8\,\mathrm{m}

Solution:

Magnifying power of telescope,

m = \frac{f_o}{f_e}

\;\text{ here }\;\; f_{o}=15\,\mathrm{m}, f_e=1.0\,\mathrm{cm}=0.01\,\mathrm{m}

\therefore\; m = \frac{15}{0.01} = 1500.

Let

D

be diameter of moon,

d

be diameter of image of moon formed by objective and

r

the distance of moon from objective lens then from figure,

\frac{D}{r} = \frac{d}{f_o}

\Rightarrow \;\; d = \frac{D}{r} \cdot f_{o}

= \frac{3.48 \times 10^6}{3.8 \times 10^8} \times 15\,\mathrm{m}

= 0.137\,\mathrm{m} = 13.7\,\mathrm{cm}.

Go to Question:

Prev Question

Next Question