JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 10)
A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 $$\mathop A\limits^ \circ $$ and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :
3 mm
9 mm
4.5 mm
1.5 mm
Explanation
The slit width is a = 0.1 mm = 10$$-$$4 m.
The wavelength of the light is $$\lambda$$ = 6000 $$\times$$ 10$$-$$10 = 6 $$\times$$ 10$$-$$7.
The distance from the slit to diffraction bands is D = 0.5 m.
We calculate the third dark band from the central band as follows:
$$A\sin \theta = 3\lambda \Rightarrow \sin \theta = {{3\lambda } \over a} = {x \over D}$$
$$ \Rightarrow x = {{3\lambda D} \over a} = {{3 \times 6 \times {{10}^{ - 7}} \times 0.5} \over {{{10}^{ - 4}}}} = 9$$ mm
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