Question

A ray of light travelling in air falls on the surface of a transparent material at an angle of 45° to the normal. It bends by 15° after refraction. Find the refractive index of the material.​

Answer 1

Explanation:

refractive index = √2

by Snell law

refractive angle will be 30

Answer 2

➭ Question :-

$\longrightarrow$A ray of light travelling in air falls on the surface of a transparent material at an angle of 45° to the normal. It bends by 15° after refraction. Find the refractive index of the material.

➭ Given:-

$\longrightarrow$Angle of incident = 45

$\longrightarrow$Angle of refraction = 15

➭ To Find:-

$\longrightarrow$The refractive index of the material

➭ Solution:-

As we know,

Refractive index of air is 1

$\longrightarrow \rm RI \: of \: the \: transparent \: medium = \\ \\ \longrightarrow \rm 1 + \frac{angle \: of \: incident}{angle \: of \: refrcation} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm on \: substituting \: the \: value \: we \: get \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \rm 1 \: + \cancel \frac{15}{45} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \rm 1 + 0.333 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \boxed {\rm 1.333} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence, the refractive index is 1.333

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MoRe to know.....

• Refractive Index (Index of Refraction) is a value calculated from the ratio of the speed of light in a vacuum to that in a second medium of greater density.
• The refractive index variable is most commonly symbolized by the letter n or n' in descriptive text and mathematical equations.