A beam of light is incident vertically on a glass slab of thickness 1 cm, and refractive index 1.5. A fraction

Question

A beam of light is incident vertically on a glass slab of thickness 1 cm, and refractive index 1.5. A fraction A is reflected from the front surface while another fraction B enters the slab and emerges after reflection from the back surface. The time delay between them is:

Accepted Answer

1. **Speed of Light in Medium:** The speed of light ($v$) in a medium of refractive index $n$ is given by $v = c/n$, where $c$ is the speed of light in vacuum ($3 	imes 10^8$ m/s) .
   $$ v = \frac{3 	imes 10^8}{1.5} = 2 	imes 10^8 	ext{ m/s} $$
2. **Path Difference:** Ray A reflects from the front surface. Ray B travels through the slab thickness ($d$), reflects from the back surface, and travels back through the slab to emerge from the front. The total extra distance covered by Ray B is $2d$.
   Given thickness $d = 1 	ext{ cm} = 1 	imes 10^{-2} 	ext{ m}$.
   $$ 	ext{Total Distance} = 2d = 2 	imes 10^{-2} 	ext{ m} $$
3. **Time Delay Calculation:** The time delay ($\Delta t$) is the time taken to travel the extra distance in the medium.
   $$ \Delta t = \frac{	ext{Total Distance}}{	ext{Speed}} = \frac{2 	imes 10^{-2}}{2 	imes 10^8} $$
   $$ \Delta t = 10^{-10} 	ext{ s} $$
4. **Discrepancy Note:** The calculated answer is $10^{-10}$ s (Option A). The provided probable answer ($5 	imes 10^{-11}$ s) corresponds to a distance of 1 cm (one-way travel) or a thickness of 0.5 cm, suggesting a potential error in the question text's given thickness or the provided answer key.

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