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When light rays are incident on a thick ...

When light rays are incident on a thick transparent slab then these rays are partly reflected and partly refracted into the slab. At a certain angle of incidence `theta`, it is found that reflected and refracted rays are mutually perpendicular. What is refractive index of material ?

A

`sin theta`

B

`cos theta`

C

`tan theta`

D

`cot theta`

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the refractive index of the material of the thick transparent slab when the reflected and refracted rays are mutually perpendicular at a certain angle of incidence \( \theta \). ### Step-by-Step Solution: 1. **Understanding the Geometry**: - When light hits the slab, part of it is reflected and part is refracted. We are given that the reflected ray and the refracted ray are perpendicular to each other. This means that if the angle of incidence is \( \theta \), the angle of reflection is also \( \theta \) (according to the law of reflection). 2. **Setting Up Angles**: - Let the angle of refraction be \( r \). Since the reflected and refracted rays are perpendicular, we have: \[ \theta + r = 90^\circ \] - This implies: \[ r = 90^\circ - \theta \] 3. **Applying Snell's Law**: - According to Snell's Law, we have: \[ n_1 \sin(\theta) = n_2 \sin(r) \] - Here, \( n_1 = 1 \) (the refractive index of air), \( n_2 = \nu \) (the refractive index of the slab), and \( r = 90^\circ - \theta \). - Therefore, we can rewrite Snell's Law as: \[ 1 \cdot \sin(\theta) = \nu \cdot \sin(90^\circ - \theta) \] 4. **Using the Identity for Sine**: - We know that \( \sin(90^\circ - \theta) = \cos(\theta) \). Substituting this into the equation gives: \[ \sin(\theta) = \nu \cdot \cos(\theta) \] 5. **Rearranging the Equation**: - Rearranging the equation to solve for \( \nu \): \[ \nu = \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \] 6. **Conclusion**: - Thus, the refractive index \( \nu \) of the material of the slab is: \[ \nu = \tan(\theta) \] ### Final Answer: The refractive index of the material is \( \nu = \tan(\theta) \).

To solve the problem, we need to determine the refractive index of the material of the thick transparent slab when the reflected and refracted rays are mutually perpendicular at a certain angle of incidence \( \theta \). ### Step-by-Step Solution: 1. **Understanding the Geometry**: - When light hits the slab, part of it is reflected and part is refracted. We are given that the reflected ray and the refracted ray are perpendicular to each other. This means that if the angle of incidence is \( \theta \), the angle of reflection is also \( \theta \) (according to the law of reflection). 2. **Setting Up Angles**: ...
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Knowledge Check

  • On a glass plate, a light wave is incident at an angle of 60^(@) . If the reflected and the refracted waves are mutually perpendicular, the refractive index of material is

    A
    `(sqrt(3))/(2)`
    B
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    C
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    A
    `sqrt(3)/2`
    B
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    C
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    D
    `1/sqrt(3)`
  • A ray of light strikes a material's slab at an angle of incidence 60^(@) . If the reflected and refracted rays are perpendicular to each other, the refractive index of the material is

    A
    `(1)/(sqrt(3))`
    B
    `(1)/(sqrt(2))`
    C
    `sqrt2`
    D
    `sqrt3`
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