Figure 36–35 shows the bright fringes that lie within the central diffraction envelope in two double-slit diffraction experiments using the same wavelength of light. Is (a) the slit width $a$, (b) the slit separation $d$, and (c) the ratio $d/a$in experiment B greater than, less than, or the same as those quantities in experiment A?

The width of the slit is $\text{a}$.

Slit separation is $\text{d}$.

The expression for the minima in the diffraction pattern is given as follows.

$\mathrm{asin\theta }=\mathrm{m\lambda }$ (1)

Here, $\mathrm{\lambda }$is the wavelength, is the number of fringes in the envelope, and is the angle of diffraction.

Calculate the first order minima in the diffraction pattern for experiment A,

Substitute for into equation (1).

$$\begin{gathered} a_A\sin {\theta }_A=\left(1\right)\lambda \\ {a}_A\sin {\theta }_A=\lambda \end{gathered}$$ (2)

Calculate the first order minima in the diffraction pattern for experiment B,

Substitute for into equation (1).

$$\begin{gathered} a_B\sin {\theta }_B=\left(1\right)\lambda \\ {a}_B\sin {\theta }_B=\lambda \end{gathered}$$ (3)

The first-order bright and dark fringes are closer to the central maxima. The first order minima in A are closer to the center than the first order minimum in B,${\theta }_A<{\theta }_B$

Thus,

$\sin {\theta }_A<\sin {\theta }_B$

Divide the equation (2) with (3),

$$\begin{gathered} \frac{a_A\sin {\theta }_A}{a_B\sin {\theta }_B}=\frac{\lambda }{\lambda } \\ \frac{a_A\sin {\theta }_A}{a_B\sin {\theta }_B}=1 \\ \frac{a_A}{a_B}=\frac{\sin {\theta }_B}{\sin {\theta }_A} \end{gathered}$$

Since $\sin {\theta }_A<\sin {\theta }_B$the ratio of $a_A/a_B$is greater than 1.

Therefore, ${\text{a}}_{\text{A}}>{\text{a}}_{\text{B}}$

Hence, the slit width in experiment B is less than the slid width in experiment A.

The expression for the constructive interference in the double slit experiment is given as follows.

$\mathrm{dsin\theta }=\mathrm{m\lambda }$ (4)

Let's assume the slit width is constant and slit separation is increasing between the slits. The fringes come closer and the envelop pattern remains unchanged. Therefore, more interference maxima fall in experiment B.

Hence, the slit separation is greater in experiment B than slit separation in experiment A.

Calculate the first order minim in the diffraction pattern.

Substitute 1 for into equation (1).

$a\sin \theta =\lambda$ (5)

Divide the equation (4) with (5).

$$\begin{gathered} \frac{d\sin \theta }{a\sin \theta }=\frac{m\lambda }{\lambda } \\ \frac{d}{a}=m \end{gathered}$$

From the above, the ratio of slit separation and slit width is equal to the number of fringes in the envelope.

Hence, the ratio $d/a$in experiment B is greater than the ratio of $d/a$in experiment A.