JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 6)
Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength $$\lambda $$ such that the first minima occurs directly in front of the slit (S1) ?
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$${\lambda \over {2\left( {5 - \sqrt 2 } \right)}}$$
$${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$$
$${\lambda \over {\left( {5 - \sqrt 2 } \right)}}$$
$${\lambda \over {\left( {\sqrt 5 - 2} \right)}}$$
Explanation
Path difference, S2P – S1P = $${\lambda \over 2}$$
$$ \Rightarrow $$ $$\sqrt {4{d^2} + {d^2}} - 2d = {\lambda \over 2}$$
$$ \Rightarrow $$ $$d\left( {\sqrt 5 - 2} \right) = {\lambda \over 2}$$
$$ \Rightarrow $$ d = $${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$$
$$ \Rightarrow $$ $$\sqrt {4{d^2} + {d^2}} - 2d = {\lambda \over 2}$$
$$ \Rightarrow $$ $$d\left( {\sqrt 5 - 2} \right) = {\lambda \over 2}$$
$$ \Rightarrow $$ d = $${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$$
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