Question 159: Waves Optics

Question

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.)

Accepted Answer

Given 2$$	heta $$ = 60
o
$$ \Rightarrow $$ $$	heta $$ = 30
o
We know,
$$a$$ sin $$	heta $$ = n $$\lambda $$
for first minima n = 1,
$$	herefore$$ At first minima
$$a$$ sin = $$\lambda $$
$$ \Rightarrow $$$$\,\,\,$$ 10
$$-$$6
$$ 	imes $$ sin 30
o
= $$\lambda $$
$$ \Rightarrow $$$$\,\,\,$$ $$\lambda $$ = $${{{{10}^{ - 6}}} \over 2}$$ m
Now after making a new slit,
$$	herefore$$ Fringe width,  $$\beta $$ = $${{\lambda D} \over d}$$
given, D = 50 cm and $$\beta $$ = 1 cm.
$$	herefore$$ 1 $$ 	imes $$ 10
$$-$$2
= $${{0.5 	imes {{10}^{ - 6}} 	imes 50 	imes {{10}^{ - 2}}} \over d}$$
$$ \Rightarrow $$ d = 25 $$ 	imes $$ 10
$$-$$6
m = 25 $$\mu $$m.

Answer

100 $$\mu $$m

Answer

25 $$\mu $$m

Answer

50 $$\mu $$m

Answer

75 $$\mu $$m

Waves Optics

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