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 ____________.

Given, b₁ = 3b₂<br/><br/>Here, b₁ = width of the one of the two slit and b₂ = width of the other slit.<br/><br/>As we know that,<br/><br/>Intensity, I $\propto$ (Amplitude)²<br/><br/>$\Rightarrow$ $${{{I_1}} \over {{I_2}}} = {\left( {{{{b_1}} \over {{b_2}}}} \right)^2} \Rightarrow {{{I_1}} \over {{I_2}}} = {\left( {{{3{b_2}} \over {{b_2}}}} \right)^2}$$<br/><br/>${I_1} = 9{I_2}$<br/><br/>As we know, the ratio of the minimum intensity to the maximum intensity in the interference pattern,<br/><br/>$${{{I_{\min }}} \over {{I_{\max }}}} = {\left( {{{\sqrt {{I_1}} - \sqrt {{I_2}} } \over {\sqrt {{I_1}} + \sqrt {{I_2}} }}} \right)^2}$$<br/><br/>Substituting the values in the above equations, we get<br/><br/>$${{{I_{\min }}} \over {{I_{\max }}}} = {{{{(\sqrt {9{I_2}} - \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {9{I_2}} + \sqrt {{I_2}} )}^2}}} = {\left( {{{3 - 1} \over {3 + 1}}} \right)^2}$$<br/><br/>${{{I_{\min }}} \over {{I_{\max }}}} = {1 \over 4}$<br/><br/>Comparing with, ${{{I_{\min }}} \over {{I_{\max }}}} = {x \over 4}$<br/><br/>The value of x = 1.