Question 78: Wave Optics

Question

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is $x \mathrm{~d}$. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is $9: 4$ then what is the value of $x$ ?
(Assume that the field strength varies according to the slit width.)

Accepted Answer

Let the amplitude from the slit of width $$d$$ be proportional to $$E$$ and from the slit of width $$xd$$ be proportional to $$xE$$ (with $$x>1$$, as is evident from the given options).
For two coherent waves with amplitudes $$E_1$$ and $$E_2$$, the resultant intensity when added in phase (maximum) and out of phase (minimum) is given by
$$ I_{\max} \propto (E_1+E_2)^2 \quad 	ext{and} \quad I_{\min} \propto (E_2-E_1)^2. $$
Here, setting
$$E_1 = E$$ (from the narrower slit of width $$d$$) and
$$E_2 = xE$$ (from the wider slit of width $$xd$$),
we have
$$ I_{\max} \propto (E+xE)^2 = E^2 (1+x)^2, $$
and since $$x > 1$$ the subtraction in the destructive case gives
$$ I_{\min} \propto (xE-E)^2 = E^2 (x-1)^2. $$
The ratio of maximum to minimum intensity is therefore
$$ \frac{I_{\max}}{I_{\min}} = \frac{(1+x)^2}{(x-1)^2}. $$
We are given that
$$ \frac{(1+x)^2}{(x-1)^2} = \frac{9}{4}. $$
Taking the square root of both sides (noting that all quantities are positive) leads to
$$ \frac{1+x}{x-1} = \frac{3}{2}. $$
To solve for $$x$$, cross-multiply:
$$ 2(1+x) = 3(x-1). $$
Expanding both sides:
$$ 2 + 2x = 3x - 3. $$
Rearrange the equation to isolate $$x$$:
$$ 2 + 2x + 3 = 3x \quad \Longrightarrow \quad 5 + 2x = 3x, $$
which implies
$$ x = 5. $$
Thus, the value of $$x$$ is
$$ \boxed{5}. $$

Answer

2

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3

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4

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5

Wave Optics

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