CBSE 2022 · Region 1 · Set 1 · Q11 · 3 marks
(i)Monochromatic light is incident on a surface separating two media. The frequency of the light after refraction remains unaffected but its wavelength changes. Why ?(ii)The frequency of an electromagnetic radiation is $\displaystyle 1.0 \times 10^{11} \mathrm{~Hz}$. Identify the radiation and mention its two uses.(i)Trace the path of a ray of light PQ which is incident at an angle i on one face of a glass prism of angle A. It then emerges out from the other face at an angle e. Use the ray diagram to prove that the angle through which the ray is deviated is given by $\displaystyle \angle \delta$ $\displaystyle =\angle \mathrm{i}+\angle \mathrm{e}-\angle \mathrm{A}$.(ii)What will be the minimum value of $\displaystyle \delta$ if the ray passes symmetrically through the prism ?
(i)
Monochromatic light is incident on a surface separating two media. The frequency of the light after refraction remains unaffected but its wavelength changes. Why ?
(ii)
The frequency of an electromagnetic radiation is $\displaystyle 1.0 \times 10^{11} \mathrm{~Hz}$. Identify the radiation and mention its two uses.
(i)
Trace the path of a ray of light PQ which is incident at an angle i on one face of a glass prism of angle A. It then emerges out from the other face at an angle e. Use the ray diagram to prove that the angle through which the ray is deviated is given by $\displaystyle \angle \delta$ $\displaystyle =\angle \mathrm{i}+\angle \mathrm{e}-\angle \mathrm{A}$.
(ii)
What will be the minimum value of $\displaystyle \delta$ if the ray passes symmetrically through the prism ?
Marking-scheme solution
(i)
Refraction arises through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators which take up the frequency of the external agency causing forced oscillations. Thus the frequency of refracted light equals the frequency of incident light.
Alternatively
Frequency is the characteristic of the source of light. So it remains unaffected. But $\displaystyle \lambda$ depends on refractive index ( $\displaystyle \mu$ ) of the medium as -
$$\lambda_{m}=\frac{\lambda_{o}}{\mu}
$$ii) Infrared/ Microwaves/ Radio waves
Uses of Infrared rays
Remote control
Green house effect
Photography in foggy condition
To reveal secret writings
Infrared lamps
Uses of Microwaves
Radar System
Geostationary satellite
Microwave ovens
Uses of Radiowaves
TV transmission
Radio broadcast
Mobile transmission
OR
Diagram
$\displaystyle \delta=\left(i-r_{1}\right)+\left(e-r_{2}\right)$
$\displaystyle \angle \mathrm{Q}=\angle \mathrm{R}=90^{\circ} \quad \therefore \angle \mathrm{A}+\angle \mathrm{O}=180^{\circ}$
In $\displaystyle \triangle \mathrm{QOR}$
$$\begin{equation*}
O+r_{$\displaystyle 1$}+r_{$\displaystyle 2$}=$\displaystyle 180$^{\circ} \tag{2}
\end{equation*}
$$Comparing ($\displaystyle 1$) and
$$\therefore A=r_{$\displaystyle 1$}+r_{$\displaystyle 2$}
$$$\therefore \delta=(i+e)-A$
Ray Optics and Optical InstrumentsRefraction through a PrismUnderstandshort_answermedium
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.