JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 8)
In Young’s double slit experiment, the distance between slits and the screen is
1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d0 between the slits. If the angular resolution of the eye is $$({{{1}} \over {60}})^o$$, the value of d0 is close to :
1 mm
2 mm
4 mm
3 mm
Explanation
We know,
Fringe width, $$\beta $$ = $${{\lambda D} \over {{d_0}}}$$
_9th_April_Morning_Slot_en_8_1.png)
From image,
$$\theta $$ = $${\beta \over D}$$
$$ \Rightarrow $$ $$\theta $$ = $${\lambda \over {{d_0}}}$$
$$ \Rightarrow $$ d0 = $${\lambda \over \theta }$$
= $$ {{600 \times {{10}^{ - 9}}} \over {{1 \over {60}} \times {\pi \over {180}}}}$$
= 2.06 $$ \times $$ 10$$-$$3
$$ \simeq $$ 2 mm
Fringe width, $$\beta $$ = $${{\lambda D} \over {{d_0}}}$$
From image,
$$\theta $$ = $${\beta \over D}$$
$$ \Rightarrow $$ $$\theta $$ = $${\lambda \over {{d_0}}}$$
$$ \Rightarrow $$ d0 = $${\lambda \over \theta }$$
= $$ {{600 \times {{10}^{ - 9}}} \over {{1 \over {60}} \times {\pi \over {180}}}}$$
= 2.06 $$ \times $$ 10$$-$$3
$$ \simeq $$ 2 mm
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