Explanation
The incident ray $$1$$ will be reflected at an equal angle as reflected ray $$2.$$
By laws of reflection, $$i=r=60^o$$
The refracted ray $$3$$ is perpendicular to reflected ray $$2.$$
So by geometry, the angle of refraction $$r'=30^o$$
Now by Snell's law,
$$\mu_{air} \ sin \ i= \mu_{glass} \ sin \ r'$$
$$(1) \times sin \ 60^o= \mu_{glass} \times sin \ 30^o$$
$$\dfrac{\sqrt3}{2}=\mu_{glass} \times \dfrac{1}{2}$$
$$\mu_{glass}=\sqrt3$$
Option A is the answer.
From the figure we can see that the right angled prism needs to be rotated by $${45^\circ }$$.
Hence the correct answer is option (A).
The correct option is (C).
Given,
The focal length of the objective lens $${f_o} = 100\;cm$$
The focal lens of the eyepiece $${f_e} = 2\;cm$$
The magnifying power of a telescope can be given as the ratio of the focal length of the objective lens$${f_o}$$ to the focal length of the eyepiece $${f_e}$$ fo the telescope, hence
$$M = \frac{{{f_o}}}{{{f_e}}}$$
$$M = \frac{{100}}{2}$$
$$M = 50$$ .
The eye lens is an elastic muscular lens that controls the formation of an image on the retina. It is connected by ciliary muscles which control the shape of the eye lens. When the light coming from a far object the lens got stretched by ciliary muscles results in a decrease in the thickness of the lens and when light coming from nearby objects the thickness of the lens increases.
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