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Question: A beaker containing liquid is placed on a table underneath a microscope with can be moved along a ve...

A beaker containing liquid is placed on a table underneath a microscope with can be moved along a vertical scale. The microscope is focussed, through the liquid into a mark on the table when the reading on the scale is a. It is next focussed on the upper surface of the liquid and the reading be b. More the liquid is added and the observations are repeated, the corresponding readings are C and d. The refractive index of the liquid is

A

d−bd−c−b+a\frac{d - b}{d - c - b + a}

B

b−dd−c−b+a\frac{b - d}{d - c - b + a}

C

d−c−b+ad−b\frac{d - c - b + a}{d - b}

D

d−ba+b−c−d\frac{d - b}{a + b - c - d}

Answer

d−bd−c−b+a\frac{d - b}{d - c - b + a}

Explanation

Solution

The real depth = (R.I.) apparent depth

⇒ In first case, the real depth h1 = n(b-a)

Similarly in the second case, the real depth h2 = n (d-c)

Since, h2 > h1 the difference of real depths = h2 - h1 = n (d-c-b+a).

Since the liquid is added in second case, h2 – h1 = d-b

⇒ n = d−bd−c−b+a\frac{d - b}{d - c - b + a}