JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 16)
A Young's double-slit experiment is performed using monochromatic light of wavelength $$\lambda $$. The
intensity of light at a point on the screen, where the path difference is $$\lambda $$, is K units. The intensity
of light at a point where the path difference is $${\lambda \over 6}$$ is given by $${{nK} \over {12}}$$, where n is an integer. The value
of n is __________.
Answer
9
Explanation
From 1st case,
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda } \times \lambda $$ = 2$$\pi $$
$$ \therefore $$ Inet = 4Icos2 $${{\Delta \phi } \over 2}$$ = 4I = K (Given)
From 2nd case,
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda } \times {\lambda \over 6}$$ = $${\pi \over 3}$$
$$ \therefore $$ Inet = 4I cos2 $${{\Delta \phi } \over 2}$$
= 4I $$ \times $$ $${3 \over 4}$$ = $${3 \over 4}K$$
$$ \therefore $$ According to question,
$${{nK} \over {12}}$$ = $${3 \over 4}K$$
$$ \Rightarrow $$ n = 9
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda } \times \lambda $$ = 2$$\pi $$
$$ \therefore $$ Inet = 4Icos2 $${{\Delta \phi } \over 2}$$ = 4I = K (Given)
From 2nd case,
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda } \times {\lambda \over 6}$$ = $${\pi \over 3}$$
$$ \therefore $$ Inet = 4I cos2 $${{\Delta \phi } \over 2}$$
= 4I $$ \times $$ $${3 \over 4}$$ = $${3 \over 4}K$$
$$ \therefore $$ According to question,
$${{nK} \over {12}}$$ = $${3 \over 4}K$$
$$ \Rightarrow $$ n = 9
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