Question:
A ray of light from a denser medium strikes a rarer medium at an angle of incidence i (see figure). The reflected and refracted rays make an angle of $90^\circ$ with each other. The angles of reflection and refraction are r and r'. The critical angle is
A ray of light from a denser medium strikes a rarer medium at an angle of incidence i (see figure). The reflected and refracted rays make an angle of $90^\circ$ with each other. The angles of reflection and refraction are r and r'. The critical angle is
Updated On: Apr 11, 2024
- $sin^{-1}$(tan r)
- $sin^{-1}$(cot i)
- $sin^{-1}$(tan r')
- $tan^{-1}$(tan i)
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The Correct Option is A
Solution and Explanation
$r+r'+990^\circ=180^\circ$
$\therefore \hspace15mm r'=90^\circ-r$
$Futher, \hspace15mm i=r$
Applying Snell's law $\mu_D sin\, i=\mu_R$ sin r'
or $\mu_D sin\, r=\mu_R sin(90^\circ-r)=\mu^R cos r$
$\therefore\, \, \, \, \, \frac{mu_R}{mu_D}=tan\, r, \theta_C=sin^{-1}\Bigg(\frac{\mu_R}{\mu_D}\Bigg)=sin^{-1}(tan\, r)$
$\therefore$ Correct option is (a).
$\therefore \hspace15mm r'=90^\circ-r$
$Futher, \hspace15mm i=r$
Applying Snell's law $\mu_D sin\, i=\mu_R$ sin r'
or $\mu_D sin\, r=\mu_R sin(90^\circ-r)=\mu^R cos r$
$\therefore\, \, \, \, \, \frac{mu_R}{mu_D}=tan\, r, \theta_C=sin^{-1}\Bigg(\frac{\mu_R}{\mu_D}\Bigg)=sin^{-1}(tan\, r)$
$\therefore$ Correct option is (a).
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Concepts Used:
Reflection of Light
When a light ray falls on any object, it is bounced back from the object. This process is known as the Reflection of light. The light reflected from the object falls into our eyes, making the object visible to us. All the things we see around us are because of reflection.
The reflection of light depends on the type of object. A polished or smooth surface reflects most of the light falling on it, while a rough surface absorbs some amount of light and reflects back the rest of the light. The direction of reflected rays depends upon the surface of the object.