Question

When the light ray illustrated in the figure below passes through the glass block of index of refraction n = 1.50, it is shifted laterally by the distance d. (Let L = 2.05 cm and ? = 31.0°.) (a) Find the value of d. .428 cm (b) Find the time interval required for the light to pass through the glass block. ps

          When the light ray illustrated in the figure below passes through the glass block of index of refraction n = 1.50, it is shifted laterally by the distance d. (Let L = 2.05 cm and ? = 31.0°.)
(a) Find the value of d.
.428 cm
(b) Find the time interval required for the light to pass through the glass block.
ps

$When the light ray illustrated in the figure below passes through the glass block of index of refraction n = 1.50, it is shifted laterally by the distance d. (Let L = 2.05 cm and ? = 31.0°.) (a) Find the value of d. .428 cm (b) Find the time interval required for the light to pass through the glass block. ps$

Added by Alex B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prabhu Ramji

08/07/2023

Step 1

First, we need to find the angle of refraction (r) using Snell's Law: n1 * sin(θ1) = n2 * sin(θ2) where n1 = 1 (air), θ1 = 31.08°, and n2 = 1.50 (glass). 1 * sin(31.08°) = 1.50 * sin(θ2) Solving for θ2, we get: θ2 = arcsin(1 * sin(31.08°) / 1.50) θ2 ≈ 19.47° Show more…

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When the light ray illustrated in the figure below passes through the glass block of index of refraction n = 1.50, it is shifted laterally by the distance d. (Let L = 2.05 cm and θ = 31.0°.) (a) Find the value of d. (b) Find the time interval required for the light to pass through the glass block.