JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 2)
Orange light of wavelength 6000 $$ \times $$ 10–10 m illuminates a single
slit of width 0.6 $$ \times $$ 10–4 m. The maximum possible number of diffraction minima produced on both sides of the central maximum is ___________.
slit of width 0.6 $$ \times $$ 10–4 m. The maximum possible number of diffraction minima produced on both sides of the central maximum is ___________.
Answer
200
Explanation
For minima
$$d\,\sin \theta = n\lambda $$
or $$\sin \theta = {{n\lambda } \over d}$$
$$ \because $$ maximum value of sin$$\theta $$ is 1
$$ \therefore $$ $${{n\lambda } \over d} \le 1$$
$$n \le {d \over \lambda }$$
$$n \le {{0.6 \times {{10}^{ - 4}}} \over {6000 \times {{10}^{ - 10}}}}$$
$$n \le 100$$
For both sides 100 + 100 = 200
$$d\,\sin \theta = n\lambda $$
or $$\sin \theta = {{n\lambda } \over d}$$
$$ \because $$ maximum value of sin$$\theta $$ is 1
$$ \therefore $$ $${{n\lambda } \over d} \le 1$$
$$n \le {d \over \lambda }$$
$$n \le {{0.6 \times {{10}^{ - 4}}} \over {6000 \times {{10}^{ - 10}}}}$$
$$n \le 100$$
For both sides 100 + 100 = 200
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