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NEET PHYSICSMedium

An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm5\text{ cm} deep when viewed from one surface and 3 cm3\text{ cm} deep when viewed from the opposite face. The thickness (in cm) of the slab is:

A

8

B

10

C

12

D

16

Step-by-Step Solution

  1. Formula for Apparent Depth: When an object inside a denser medium (refractive index μ\mu) is viewed from a rarer medium (air), the relationship between real depth (dreald_{real}) and apparent depth (dappd_{app}) is given by: dreal=μ×dappd_{real} = \mu \times d_{app}

  2. Calculation from Surface 1: Apparent depth, dapp1=5 cmd_{app1} = 5\text{ cm}. Real depth from surface 1, x=μ×dapp1=1.5×5=7.5 cmx = \mu \times d_{app1} = 1.5 \times 5 = 7.5\text{ cm}.

  3. Calculation from Surface 2: Apparent depth, dapp2=3 cmd_{app2} = 3\text{ cm}. Real depth from surface 2, y=μ×dapp2=1.5×3=4.5 cmy = \mu \times d_{app2} = 1.5 \times 3 = 4.5\text{ cm}.

  4. Total Thickness: The total thickness of the slab (tt) is the sum of the real depths of the bubble from both surfaces. t=x+y=7.5 cm+4.5 cm=12 cmt = x + y = 7.5\text{ cm} + 4.5\text{ cm} = 12\text{ cm}.

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