JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 21)
The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x : 4 where x is ____________.
Answer
1
Explanation
Given, b1 = 3b2
Here, b1 = width of the one of the two slit and b2 = width of the other slit.
As we know that,
Intensity, I $$\propto$$ (Amplitude)2
$$\Rightarrow$$ $${{{I_1}} \over {{I_2}}} = {\left( {{{{b_1}} \over {{b_2}}}} \right)^2} \Rightarrow {{{I_1}} \over {{I_2}}} = {\left( {{{3{b_2}} \over {{b_2}}}} \right)^2}$$
$${I_1} = 9{I_2}$$
As we know, the ratio of the minimum intensity to the maximum intensity in the interference pattern,
$${{{I_{\min }}} \over {{I_{\max }}}} = {\left( {{{\sqrt {{I_1}} - \sqrt {{I_2}} } \over {\sqrt {{I_1}} + \sqrt {{I_2}} }}} \right)^2}$$
Substituting the values in the above equations, we get
$${{{I_{\min }}} \over {{I_{\max }}}} = {{{{(\sqrt {9{I_2}} - \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {9{I_2}} + \sqrt {{I_2}} )}^2}}} = {\left( {{{3 - 1} \over {3 + 1}}} \right)^2}$$
$${{{I_{\min }}} \over {{I_{\max }}}} = {1 \over 4}$$
Comparing with, $${{{I_{\min }}} \over {{I_{\max }}}} = {x \over 4}$$
The value of x = 1.
Here, b1 = width of the one of the two slit and b2 = width of the other slit.
As we know that,
Intensity, I $$\propto$$ (Amplitude)2
$$\Rightarrow$$ $${{{I_1}} \over {{I_2}}} = {\left( {{{{b_1}} \over {{b_2}}}} \right)^2} \Rightarrow {{{I_1}} \over {{I_2}}} = {\left( {{{3{b_2}} \over {{b_2}}}} \right)^2}$$
$${I_1} = 9{I_2}$$
As we know, the ratio of the minimum intensity to the maximum intensity in the interference pattern,
$${{{I_{\min }}} \over {{I_{\max }}}} = {\left( {{{\sqrt {{I_1}} - \sqrt {{I_2}} } \over {\sqrt {{I_1}} + \sqrt {{I_2}} }}} \right)^2}$$
Substituting the values in the above equations, we get
$${{{I_{\min }}} \over {{I_{\max }}}} = {{{{(\sqrt {9{I_2}} - \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {9{I_2}} + \sqrt {{I_2}} )}^2}}} = {\left( {{{3 - 1} \over {3 + 1}}} \right)^2}$$
$${{{I_{\min }}} \over {{I_{\max }}}} = {1 \over 4}$$
Comparing with, $${{{I_{\min }}} \over {{I_{\max }}}} = {x \over 4}$$
The value of x = 1.
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