JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 12)
Light of wavelength $$550$$ $$nm$$ falls normally on a slit of width $$22.0 \times {10^{ - 5}}$$ $$cm.$$ The angular position of the second minima from the central maximum will (in radians) :
$${\pi \over {12}}$$
$${\pi \over 8}$$
$${\pi \over 6}$$
$${\pi \over 4}$$
Explanation
Angular position of nth minima from central maxima,
sin $$\theta $$ = $${{n\lambda } \over a}$$
here n = 2
$$\therefore\,\,\,\,$$ sin $$\theta $$ = $${{2 \times 550 \times {{10}^{ - 9}}} \over {22 \times {{10}^{ - 5}}}}$$ = $${1 \over 2}$$
$$\therefore\,\,\,$$ $$\theta $$ = $${\pi \over 6}$$ rad.
sin $$\theta $$ = $${{n\lambda } \over a}$$
here n = 2
$$\therefore\,\,\,\,$$ sin $$\theta $$ = $${{2 \times 550 \times {{10}^{ - 9}}} \over {22 \times {{10}^{ - 5}}}}$$ = $${1 \over 2}$$
$$\therefore\,\,\,$$ $$\theta $$ = $${\pi \over 6}$$ rad.
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