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If a ray of light in a denser medium str...

If a ray of light in a denser medium strikes a rarer medium at an angle of incidence i, the angles of reflection and refraction are respectively, r and r' If the reflected and refraction rays are at right angles to each other, the critical angle for the given pair of media is

A

`sin^(-1)(tanr')`

B

`sin^(-1)(tanr)`

C

`tan^(-1)(sini)`

D

`cot^(-1)(tani)`

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The correct Answer is:
To solve the problem, we need to find the critical angle (θc) for a ray of light transitioning from a denser medium to a rarer medium, given that the reflected and refracted rays are at right angles to each other. ### Step-by-Step Solution: 1. **Understanding the Situation**: - We have a ray of light moving from a denser medium (n1) to a rarer medium (n2). - The angle of incidence is denoted as \(i\), the angle of reflection as \(r\), and the angle of refraction as \(r'\). - According to the problem, the reflected ray and the refracted ray are at right angles to each other, meaning \(r + r' = 90^\circ\). 2. **Using Snell's Law**: - Snell's Law states that: \[ n_1 \sin(i) = n_2 \sin(r') \] - Since we are moving from a denser medium (n1) to a rarer medium (n2), we can express \(n_2\) in terms of \(n_1\): \[ n_2 = \frac{n_1}{n} \] - Rearranging gives us: \[ \sin(i) = \frac{n_2}{n_1} \sin(r') \] 3. **Relating Angles**: - Since \(r' = 90^\circ - r\), we can substitute this into our equation: \[ \sin(i) = \frac{n_2}{n_1} \sin(90^\circ - r) = \frac{n_2}{n_1} \cos(r) \] 4. **Expressing n2 in terms of n1**: - From Snell's Law, we can express \(n_2\) as: \[ n_2 = n_1 \sin(i) / \cos(r) \] - This can be simplified to: \[ n_2 = n_1 \tan(i) \] 5. **Finding the Critical Angle**: - The critical angle \(θ_c\) occurs when \(r' = 90^\circ\). Thus: \[ n_1 \sin(θ_c) = n_2 \] - Substituting \(n_2\) gives: \[ n_1 \sin(θ_c) = n_1 \tan(i) \] - Dividing both sides by \(n_1\) (assuming \(n_1 \neq 0\)): \[ \sin(θ_c) = \tan(i) \] 6. **Final Expression for Critical Angle**: - The critical angle can be expressed as: \[ θ_c = \sin^{-1}(\tan(i)) \] ### Conclusion: Thus, the critical angle for the given pair of media is: \[ θ_c = \sin^{-1}(\tan(i)) \]

To solve the problem, we need to find the critical angle (θc) for a ray of light transitioning from a denser medium to a rarer medium, given that the reflected and refracted rays are at right angles to each other. ### Step-by-Step Solution: 1. **Understanding the Situation**: - We have a ray of light moving from a denser medium (n1) to a rarer medium (n2). - The angle of incidence is denoted as \(i\), the angle of reflection as \(r\), and the angle of refraction as \(r'\). - According to the problem, the reflected ray and the refracted ray are at right angles to each other, meaning \(r + r' = 90^\circ\). ...
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DC PANDEY ENGLISH-RAY OPTICS-Exercise
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  4. A thin prism P with angle 4^(@) and made from glass of refractive inde...

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  5. A convex lens produces an image of a real object on a screen with a ma...

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  8. A point object is moving with a speed v before an arrangement of two m...

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  9. If a ray of light in a denser medium strikes a rarer medium at an angl...

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  10. A ray PQ incident on the refracting face BA is refracted in the prism ...

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  11. The xz plane separates two media A and B with refractive indices mu(1)...

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  12. A thin lens made of glass of refractive index mu = 1.5 has a focal len...

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  13. A ray of light is incident on a surface of glass slab at an angle 45^@...

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