JEE MAIN - Physics (2018 (Offline) - No. 2)
The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is
1 $$\mu $$m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it,
Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed
fringe width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)
100 $$\mu $$m
25 $$\mu $$m
50 $$\mu $$m
75 $$\mu $$m
Explanation
_en_2_1.png)
Given 2$$\theta $$ = 60o
$$ \Rightarrow $$ $$\theta $$ = 30o
We know,
$$a$$ sin $$\theta $$ = n $$\lambda $$
for first minima n = 1,
$$\therefore$$ At first minima
$$a$$ sin = $$\lambda $$
$$ \Rightarrow $$$$\,\,\,$$ 10$$-$$6 $$ \times $$ sin 30o = $$\lambda $$
$$ \Rightarrow $$$$\,\,\,$$ $$\lambda $$ = $${{{{10}^{ - 6}}} \over 2}$$ m
Now after making a new slit,
$$\therefore$$ Fringe width, $$\beta $$ = $${{\lambda D} \over d}$$
given, D = 50 cm and $$\beta $$ = 1 cm.
$$\therefore$$ 1 $$ \times $$ 10$$-$$2 = $${{0.5 \times {{10}^{ - 6}} \times 50 \times {{10}^{ - 2}}} \over d}$$
$$ \Rightarrow $$ d = 25 $$ \times $$ 10$$-$$6 m = 25 $$\mu $$m.
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