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A ray of light falls normally (as shown) on a right angled prism whose $μ = \sqrt{1.5}$ and 45°. After passing through prism, it falls on a concave mirror which is at a distance of 27.46cm from the point of emergence of prism. The net deviation of the ray is $n \times 90 °$ , (if focal length of the mirror is 10cm) then value of n is

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answer is 2.

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Detailed Solution

$B A D = 75 ° \tan 75 ° = \frac{B D}{A B} \Rightarrow B D = 2 (2 + \sqrt{3}) = 7.46 c m$

Hence DP = 20 cm i.e., the ray emerging from the prism passes through the centre of curvature of the mirror.

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