CBSE 2024 · Region 2 · Set 1 · Q30 · 4 marks
When a ray of light propagates from a denser medium to a rarer medium, it bends away from the normal. When the incident angle is increased, the refracted ray deviates more from the normal. For a particular angle of incidence in the denser medium, the refracted ray just grazes the interface of the two surfaces. This angle of incidence is called the critical angle for the pair of media involved.(i)For a ray incident at the critical angle, the angle of reflection is :(A)$\displaystyle 0^{\circ}$(B)$\displaystyle <90^{\circ}$(C)$\displaystyle >90^{\circ}$(D)$\displaystyle 90^{\circ}$(ii)A ray of light of wavelength $\displaystyle 600$ nm is incident in water $\displaystyle \left(\mathrm{n}=\frac{4}{3}\right)$ on the water-air interface at an angle less than the critical angle. The wavelength associated with the refracted ray is :(A)$\displaystyle 400$ nm(B)$\displaystyle 450$ nm(C)$\displaystyle 600$ nm(D)$\displaystyle 800$ nm(a)The interface AB between the two media A and B is shown in the figure. In the denser medium A , the incident ray PQ makes an angle of $\displaystyle 30^{\circ}$ with the horizontal. The refracted ray is parallel to the interface. The refractive index of medium B w.r.t. medium A is :
(A)$\displaystyle \frac{\sqrt{3}}{2}$(B)$\displaystyle \frac{\sqrt{5}}{2}$(C)$\displaystyle \frac{4}{\sqrt{3}}$(D)$\displaystyle \frac{2}{\sqrt{3}}$Two media A and B are separated by a plane boundary. The speed of light in medium A and B is $\displaystyle 2 \times 10^{8} \mathrm{~ms}^{-1}$ and $\displaystyle 2.5 \times 10^{8} \mathrm{~ms}^{-1}$ respectively. The critical angle for a ray of light going from medium A to medium B is :(A)$\displaystyle \sin ^{-1} \frac{1}{2}$(B)$\displaystyle \sin ^{-1} \frac{4}{5}$(C)$\displaystyle \sin ^{-1} \frac{3}{5}$(D)$\displaystyle \sin ^{-1} \frac{2}{5}$(iv)The figure shows the path of a light ray through a triangular prism. In this phenomenon, the angle $\displaystyle \theta$ is given by :
(A)$\displaystyle \sin ^{-1} \sqrt{\mathrm{n}^{2}-1}$(B)$\displaystyle \sin ^{-1}\left(\mathrm{n}^{2}-1\right)$(C)$\displaystyle \sin ^{-1}\left[\frac{1}{\sqrt{\mathrm{n}^{2}-1}}\right]$(D)$\displaystyle \sin ^{-1}\left[\frac{1}{\left(\mathrm{n}^{2}-1\right)}\right]$
When a ray of light propagates from a denser medium to a rarer medium, it bends away from the normal. When the incident angle is increased, the refracted ray deviates more from the normal. For a particular angle of incidence in the denser medium, the refracted ray just grazes the interface of the two surfaces. This angle of incidence is called the critical angle for the pair of media involved.
(i)
For a ray incident at the critical angle, the angle of reflection is :
(A)
$\displaystyle 0^{\circ}$
(B)
$\displaystyle <90^{\circ}$
(C)
$\displaystyle >90^{\circ}$
(D)
$\displaystyle 90^{\circ}$
(ii)
A ray of light of wavelength $\displaystyle 600$ nm is incident in water $\displaystyle \left(\mathrm{n}=\frac{4}{3}\right)$ on the water-air interface at an angle less than the critical angle. The wavelength associated with the refracted ray is :
(A)
$\displaystyle 400$ nm
(B)
$\displaystyle 450$ nm
(C)
$\displaystyle 600$ nm
(D)
$\displaystyle 800$ nm
(a)
The interface AB between the two media A and B is shown in the figure. In the denser medium A , the incident ray PQ makes an angle of $\displaystyle 30^{\circ}$ with the horizontal. The refracted ray is parallel to the interface. The refractive index of medium B w.r.t. medium A is :
(A)
$\displaystyle \frac{\sqrt{3}}{2}$
(B)
$\displaystyle \frac{\sqrt{5}}{2}$
(C)
$\displaystyle \frac{4}{\sqrt{3}}$
(D)
$\displaystyle \frac{2}{\sqrt{3}}$
Two media A and B are separated by a plane boundary. The speed of light in medium A and B is $\displaystyle 2 \times 10^{8} \mathrm{~ms}^{-1}$ and $\displaystyle 2.5 \times 10^{8} \mathrm{~ms}^{-1}$ respectively. The critical angle for a ray of light going from medium A to medium B is :
(A)
$\displaystyle \sin ^{-1} \frac{1}{2}$
(B)
$\displaystyle \sin ^{-1} \frac{4}{5}$
(C)
$\displaystyle \sin ^{-1} \frac{3}{5}$
(D)
$\displaystyle \sin ^{-1} \frac{2}{5}$
(iv)
The figure shows the path of a light ray through a triangular prism. In this phenomenon, the angle $\displaystyle \theta$ is given by :
(A)
$\displaystyle \sin ^{-1} \sqrt{\mathrm{n}^{2}-1}$
(B)
$\displaystyle \sin ^{-1}\left(\mathrm{n}^{2}-1\right)$
(C)
$\displaystyle \sin ^{-1}\left[\frac{1}{\sqrt{\mathrm{n}^{2}-1}}\right]$
(D)
$\displaystyle \sin ^{-1}\left[\frac{1}{\left(\mathrm{n}^{2}-1\right)}\right]$
Marking-scheme solution
(D)
$\displaystyle 800$ nm
(a)
(A) $\displaystyle \frac{3}{2}$
(B)
$\displaystyle \sin ^{-1}\left(\frac{4}{5}\right)$
(A)
$\displaystyle \sin ^{-1} \sqrt{n^{2}-1}$
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.