JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 9)
Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be (1 light year = 9.46 $$ \times $$ 1015 m) of the order of :
106 km
108 km
1011 km
1010 km
Тайлбар
The limit of resolution of a telescope,
$$\Delta $$$$\theta $$ = $${{1.22\,\,\lambda } \over D}$$ = $${l \over R}$$
$$ \therefore $$ $$l$$ = $${{1.22\,\,\lambda R} \over D}$$
= $${{1.22 \times 6 \times {{10}^{ - 7}} \times 10 \times 9.46 \times 10{}^{15}} \over {30 \times {{10}^{ - 2}}}}$$
= 2.31 $$ \times $$ 108 km
$$\Delta $$$$\theta $$ = $${{1.22\,\,\lambda } \over D}$$ = $${l \over R}$$
$$ \therefore $$ $$l$$ = $${{1.22\,\,\lambda R} \over D}$$
= $${{1.22 \times 6 \times {{10}^{ - 7}} \times 10 \times 9.46 \times 10{}^{15}} \over {30 \times {{10}^{ - 2}}}}$$
= 2.31 $$ \times $$ 108 km