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A ray of light is incident on a transpar...

A ray of light is incident on a transparent glass slab of refractive index 1.62. If the reflected and refracted rays are mutually perpendicular, what is the angle of incidence ? [`tan^(–1) (1.62) = 58.3^(@)`]

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To solve the problem step by step, we will use the principles of optics, specifically the laws of reflection and refraction. ### Step 1: Understanding the Problem We have a ray of light incident on a glass slab with a refractive index (μ) of 1.62. The reflected ray and the refracted ray are mutually perpendicular, meaning they form a right angle (90 degrees) with each other. ### Step 2: Setting Up the Angles Let: - \( I \) = angle of incidence ...
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Knowledge Check

  • A ray of light falls on a transparent glass slab of refractive index 1.62. If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is

    A
    `tan^(-1) (1.62)`
    B
    `tan^(-1) 1/(1.62)`
    C
    `tan^(-1) (1.33)`
    D
    `tan^(-1) 1/(1.33)`

  • A ray of light falls on a transparent glass slab of refractive index 1.62. If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is

    A
    `tan^(-1) (1.62)`
    B
    `tan^(-1) 1/(1.62)`
    C
    `tan^(-1) (1.33)`
    D
    `tan^(-1) 1/(1.33)`

  • A ray of light falls on a transparent glass slab of refractive index sqrt (2) . If the reflected and refracted rays are mutually perpendicular, then the angle of incidence is

    A
    `tan^-1(sqrt(3/2))`
    B
    `sin^(-1)(sqrt(2/3))`
    C
    `tan^-1(sqrt(2))`
    D
    `sin^(-1)(sqrt(3))`

    • Similar Questions

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