JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 3)
Visible light of wavelength 6000 $$ \times $$ 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at $$\theta $$1, then $$\theta $$1, is close to :
45o
30o
25o
20o
Explanation
For 2nd minima
$$\sin \theta = {{2\lambda } \over d }$$
$$ \Rightarrow $$ $$\sin 60^\circ = {{2\lambda } \over d}$$
$$ \Rightarrow $$ $${\lambda \over d} = {{\sqrt 3 } \over 4}$$
For 1st minima
$$\sin {\theta _1} = {\lambda \over d}$$ = $${{\sqrt 3 } \over 4}$$
$$ \Rightarrow $$ $${\theta _1} = $$ 25o
$$\sin \theta = {{2\lambda } \over d }$$
$$ \Rightarrow $$ $$\sin 60^\circ = {{2\lambda } \over d}$$
$$ \Rightarrow $$ $${\lambda \over d} = {{\sqrt 3 } \over 4}$$
For 1st minima
$$\sin {\theta _1} = {\lambda \over d}$$ = $${{\sqrt 3 } \over 4}$$
$$ \Rightarrow $$ $${\theta _1} = $$ 25o
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