Consider three media P, Q and R with refractive indices 1, 1.25, and 1.5, respectively. The medium Q having a thickness of 5 cm is placed between extended media P and R as shown in the figure. An object O is placed at the center of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is h_1. For similar observation from medium R, the apparent depth is h_2. The value of |h_1-h_2|, in cm, is :
C. 1
The object O is placed at the center of medium Q, which has a thickness of 5 cm . Therefore, the actual depth of the object from both the P-Q interface and the Q-R interface is: d = 5 2 = 2.5 cm The formula for apparent depth when viewed from a medium of refractive index n_ observer into a medium of refractive index n_ object is given by: h_ app = d n_ observer n_ object When viewed from medium P (n_P = 1), the apparent depth h_1 of the object in medium Q (n_Q = 1.25) is: h_1 = 2.5 n_P n_Q = 2.5 1 1.25 = 2.5 4 5 = 2 cm When viewed from medium R (n_R = 1.5), the apparent depth h_2 of the object in medium Q (n_Q = 1.25) is: h_2 = 2.5 n_R n_Q = 2.5 1.5 1.25 = 2.5 6/4 5/4 = 2.5 6 5 = 3 cm We need to find the value of |h_1 - h_2|: |h_1 - h_2| = |2 - 3| = |-1| = 1 cm Answer: 1
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