An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is $5 ext{ cm}$ deep when vie

Question

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

Accepted Answer

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

2.  **Calculation from Surface 1:**
    *   Apparent depth, $d_{app1} = 5	ext{ cm}$.
    *   Real depth from surface 1, $x = \mu 	imes d_{app1} = 1.5 	imes 5 = 7.5	ext{ cm}$.

3.  **Calculation from Surface 2:**
    *   Apparent depth, $d_{app2} = 3	ext{ cm}$.
    *   Real depth from surface 2, $y = \mu 	imes d_{app2} = 1.5 	imes 3 = 4.5	ext{ cm}$.

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

Step-by-Step Solution

Practice All PHYSICS Questions