Back to Directory
NEET PHYSICSMedium

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:

A

10⁻¹⁰ s

B

5 × 10⁻¹⁰ s

C

10⁻¹¹ s

D

5 × 10⁻¹¹ s

Step-by-Step Solution

  1. Speed of Light in Medium: The speed of light (vv) in a medium of refractive index nn is given by v=c/nv = c/n, where cc is the speed of light in vacuum (3×1083 \times 10^8 m/s) . v=3×1081.5=2×108 m/sv = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s}
  2. Path Difference: Ray A reflects from the front surface. Ray B travels through the slab thickness (dd), 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 2d2d. Given thickness d=1 cm=1×10−2 md = 1 \text{ cm} = 1 \times 10^{-2} \text{ m}. Total Distance=2d=2×10−2 m\text{Total Distance} = 2d = 2 \times 10^{-2} \text{ m}
  3. Time Delay Calculation: The time delay (Δt\Delta t) is the time taken to travel the extra distance in the medium. Δt=Total DistanceSpeed=2×10−22×108\Delta t = \frac{\text{Total Distance}}{\text{Speed}} = \frac{2 \times 10^{-2}}{2 \times 10^8} Δt=10−10 s\Delta t = 10^{-10} \text{ s}
  4. Discrepancy Note: The calculated answer is 10−1010^{-10} s (Option A). The provided probable answer (5×10−115 \times 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.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: PHYSICS Question for NEET | Sushrut