(C) In the first case,$\theta$ is the critical angle for the glass-air interface. According to the definition of the critical angle,$\sin \theta = \fr

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$\sin^{-1}\left(\frac{1}{\mu^2}\right)$

(C) In the first case,$\theta$ is the critical angle for the glass-air interface. According to the definition of the critical angle,$\sin \theta = \frac{1}{\mu}$. In the second case,light travels from air to glass. Using Snell's law,$\frac{\sin i}{\sin r} = \mu$,where $i = \theta$. Therefore,$\sin r = \frac{\sin \theta}{\mu}$. Substituting $\sin \theta = \frac{1}{\mu}$ into the equation,we get $\sin r = \frac{1/\mu}{\mu} = \frac{1}{\mu^2}$. Thus,the angle of refraction is $r = \sin^{-1}\left(\frac{1}{\mu^2}\right)$.

Class 12 Physics Ray Optics and Optical Instruments Critical Angle and Total Internal Reflection

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$A$ glass slab has a refractive index ' $\mu$ ' with respect to air,and the critical angle for a ray of light going from glass to air is ' $\theta$ '. If a ray of light is incident from air on the glass with an angle of incidence ' $\theta$ ',then the corresponding angle of refraction is:

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A
$\sin^{-1}\left(\frac{1}{\sqrt{\mu}}\right)$
B
$\sin^{-1}\left(\frac{1}{\mu}\right)$
C
$\sin^{-1}\left(\frac{1}{\mu^2}\right)$
D
$90^{\circ}$

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Critical Angle and Total Internal Reflection

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$A$ glass slab has a refractive index ' $\mu$ ' with respect to air,and the critical angle for a ray of light going from glass to air is ' $\theta$ '. If a ray of light is incident from air on the glass with an angle of incidence ' $\theta$ ',then the corresponding angle of refraction is:

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$A$ light ray travels from a medium with refractive index $n_1$ to another medium of refractive index $n_2$. If $n_1=2$ and $n_2=\sqrt{3}$, then find the critical angle. (in $^{\circ}$)

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$A$ beam of light consisting of red,green and blue colours is incident on a right-angled prism. The refractive index of the material of the prism for the above red,green and blue wavelengths are $1.39, 1.44$ and $1.47$ respectively.
The prism will