There is a small source of light at some depth below the surface of water

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Solution :

 [d]  $\sin \theta =\frac{3}{4}$ $\cos \theta =\frac{\sqrt{7}}{4}$ $\Omega =2\pi (1-\cos \theta )$ $=2\pi \,\,(1-\sqrt{7}/4)$ Fraction of energy transmitted$=\frac{1-\sqrt{7}/4}{2}$ $=17%$

The aperture diameter of a telescope is 5m. The separation between the moon and the earth is 4 × 105 km. With light of wavelength of 5500 $$\mathop A\limits^o$$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :
$$\theta$$ = 1.22$${\lambda \over a}$$
Distance O1O2 = ($$\theta$$)d
= (1.22$${\lambda \over a}$$)d = $${{\left( {1.22} \right)\left( {5500 \times {{10}^{ - 10}}} \right)\left( {4 \times {{10}^5}} \right) \times {{10}^3}} \over 5}$$