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Physics · 2022 · 5 marks
CBSE 2022 · Region 2 · Set 1 · Q12
The British physicist Thomas Young explained the interference of light using the principle of superposition of waves. He observed the interference…
The British physicist Thomas Young explained the interference of light using the principle of superposition of waves. He observed the interference pattern on the screen, in his experimental set-up, known now as Young's double slit experiment. The two slits $\displaystyle \mathrm{S}_{1}$ and $\displaystyle \mathrm{S}_{2}$ were illuminated by light from a slit S. The interference pattern consists of dark and bright bands of light. Such bands are called fringes. The distance between two consecutive bright and dark fringes is called fringe width.
(a)
If the screen is moved closer to the plane of slits $\displaystyle \mathrm{S}_{1}$ and $\displaystyle \mathrm{S}_{2}$, then the fringe width :
(i)
will decrease, but the intensity of bright fringe remains the same.
(ii)
will increase, but the intensity of bright fringe decreases.
(iii)
will decrease, but the intensity of bright fringe increases.
(iv)
and the intensity both remain the same.
(b)
What will happen to the pattern on the screen, when the two slits $\displaystyle \mathrm{S}_{1}$ and $\displaystyle \mathrm{S}_{2}$ are replaced by two independent but identical sources?
(i)
The intensity of pattern will increase
(ii)
The intensity of pattern will decrease
(iii)
The number of fringes will become double
(iv)
No pattern will be observed on the screen
(c)
Two sources of light are said to be coherent, when both emit light waves of:
(i)
same amplitude and have a varying phase difference.
(ii)
same wavelength and a constant phase difference.
(iii)
different wavelengths and same intensity.
(iv)
different wavelengths and a constant phase difference.
(d)
The fringe width in a Young's double slit experiment is $\displaystyle \beta$. If the whole set-up is immersed in a liquid of refractive index ' $\displaystyle \mu$ ', then the new fringe width will be :
(i)
$\displaystyle \beta$
(ii)
$\displaystyle \beta \mu$
(iii)
$\displaystyle \frac{\beta}{\mu}$
(iv)
$\displaystyle \frac{\beta}{\mu^{2}}$
(e)
The total path difference between two waves meeting at points $\displaystyle \mathrm{P}_{1}$ and $\displaystyle \mathrm{P}_{2}$ on the screen are $\displaystyle \left(\frac{3 \lambda}{2}\right)$ and $\displaystyle 2 \lambda$ respectively. Then :
(i)
bright fringes are formed at both points.
(ii)
dark fringes are formed at both points.
(iii)
a bright fringe is formed at $\displaystyle \mathrm{P}_{1}$ and a dark fringe is formed at $\displaystyle \mathrm{P}_{2}$.
(iv)
a bright fringe is formed at $\displaystyle \mathrm{P}_{2}$ and a dark fringe is formed at $\displaystyle \mathrm{P}_{1}$.
Marking-scheme solution
a) (iii)
b) (iv)
c) (ii)
d) (iii)
e) (iv)
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CBSE Class 12 Physics past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.