CBSE 2026 · Region 2 · Set 1 · Q31 · 5 marks
(a)What are coherent sources ? Why they are necessary for observing stable interference pattern ? Draw a graph showing the variation of intensity of light with the position on the screen in Young's double- slit experiment.Find the intensity of light at a point on the screen when two interfering waves of the same intensity $\displaystyle \left(\mathrm{I}_{0}\right)$ have a path difference of(i)$\displaystyle \frac{\lambda}{4}$ and (ii) $\displaystyle \frac{\lambda}{3}$.Draw a labelled ray diagram of a refracting telescope when it forms image of a distant object at infinity. Derive expression for its magnifying power.(i)In a telescope the objective has much larger aperture than the eye piece. Why ?(ii)Write two advantages of reflecting telescope over refracting telescope.
(a)
What are coherent sources ? Why they are necessary for observing stable interference pattern ? Draw a graph showing the variation of intensity of light with the position on the screen in Young's double- slit experiment.
Find the intensity of light at a point on the screen when two interfering waves of the same intensity $\displaystyle \left(\mathrm{I}_{0}\right)$ have a path difference of
(i)
$\displaystyle \frac{\lambda}{4}$ and (ii) $\displaystyle \frac{\lambda}{3}$.
Draw a labelled ray diagram of a refracting telescope when it forms image of a distant object at infinity. Derive expression for its magnifying power.
(i)
In a telescope the objective has much larger aperture than the eye piece. Why ?
(ii)
Write two advantages of reflecting telescope over refracting telescope.
Marking-scheme solution
(a)
Two sources of light emitting waves of same frequency and constant phase or zero phase difference are said to be coherent.
If the sources are incoherent, the interference pattern will change with time. Hence the positions of maxima and minima will also vary rapidly with time and we obtain a 'time averaged' intensity distribution.
Alternatively: If the two sources are coherent then the phase difference at any point will not change with time and we will have a stable interference pattern.
(b)
$\displaystyle I=4 I_{0} \cos ^{2}\left(\dfrac{\phi}{2}\right)$
$\displaystyle \Delta p=\dfrac{\lambda}{4}, \quad \phi=\dfrac{2 \pi}{\lambda} \cdot \dfrac{\lambda}{4}=\dfrac{\pi}{2}$
$\displaystyle I=4 I_{0} \cos ^{2} \dfrac{\pi}{4}=2 I_{0}$
(ii)
$\displaystyle \Delta p=\dfrac{\lambda}{3}, \quad \phi=\dfrac{2 \pi}{\lambda} \cdot \dfrac{\lambda}{3}=\dfrac{2 \pi}{3}$
$\displaystyle I=4 I_{0} \cos ^{2} \dfrac{\pi}{3}=I_{0}$
$\displaystyle m=\dfrac{\beta}{\alpha} \approx \dfrac{\tan \beta}{\tan \alpha}=\dfrac{h / f_{e}}{h / f_{o}}=\dfrac{f_{o}}{f_{e}}$
(i)
With larger aperture even the fainter objects can be observed clearly.
Alternatively: With larger aperture, the light gathering power and its resolution / resolving power increases.
Alternatively: A large aperture helps in collecting more light from a distant object, which forms a sharper / brighter image.
(ii)
There is no chromatic aberration in a mirror.
Spherical aberration in the case of parabolic reflectors is removed.
Requires less mechanical support.
Cost effective.
Brighter / sharper image is formed.
(Any two from above, or any other appropriate advantages.)
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.