A ray of light passes from glass (ng = 1.52) to water (nw = 1.33)

1.	A ray of light passes from glass (ng = 1.52) to water (nw = 1.33). What is the critical angle of incidence?
Answer» Given: n_g = 1.52, n_w = 1.33 To find: Critical angle (i_c) formula: sin i_c = \(\frac{n_2}{n_1}=\frac{n_w}{n_g}\) Calculation: From formula, i_c = sin^-1 \((\frac{1.33}{1.52})\) = sin^-1 (0.875) = 61 °2′

A ray of light passes from glass (ng = 1.52) to water (nw = 1.33). What is the critical angle of incidence?

Answer»

Given: n_g = 1.52, n_w = 1.33

To find: Critical angle (i_c)

formula: sin i_c = \(\frac{n_2}{n_1}=\frac{n_w}{n_g}\)

Calculation:

From formula,

i_c = sin^-1 \((\frac{1.33}{1.52})\) = sin^-1 (0.875) = 61 °2′

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A ray of light passes from glass (ng = 1.52) to water (nw = 1.33). What is the critical angle of incidence?

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