Question

In YDSE slits are at distance of $4\lambda$ (where $\lambda$ is wavelength of monochromatic light). If seperation between slits and screen is $D$ then ($D >> \lambda$)

• A. Total minimas observed on screen are 9
• B. Total maximas observed on screen are 10
• C. Distance of $3^{rd}$ maxima from central bright fringe is $\frac{2D}{\sqrt{7}}$
• D. Distance between first and second maxima (on same side of central maxima) is $\frac{D}{\sqrt{15}}(\sqrt{5}-1)$

Answer 1

Answer: (D)

D

Answer 2

The condition for maxima in YDSE is $d \sin \theta = n\lambda$. Given $d = 4\lambda$, we have $\sin \theta = \frac{n\lambda}{4\lambda} = \frac{n}{4}$. For observation on the screen, $|\sin \theta| \le 1$, so $|n| \le 4$. Thus, $n$ can take values $0, \pm 1, \pm 2, \pm 3, \pm 4$, giving a total of $1 + 2 \times 4 = 9$ maxima.

The condition for minima is $d \sin \theta = (2m+1)\frac{\lambda}{2}$. So, $\sin \theta = \frac{(2m+1)\lambda}{2(4\lambda)} = \frac{2m+1}{8}$. For observation on the screen, $|\sin \theta| \le 1$, so $|2m+1| \le 8$, which means $-8 \le 2m+1 \le 8$. This gives $-9 \le 2m \le 7$, or $-4.5 \le m \le 3.5$. Thus, $m$ can take values $-4, -3, -2, -1, 0, 1, 2, 3$, giving a total of $8$ minima.

The position of the $n$-th maxima is $y_n = D \tan \theta_n$. Using $\sin \theta_n = \frac{n\lambda}{d}$ and $\tan \theta_n = \frac{\sin \theta_n}{\sqrt{1-\sin^2 \theta_n}}$, we get $y_n = D \frac{n\lambda/d}{\sqrt{1-(n\lambda/d)^2}}$.

For the $3^{rd}$ maxima ($n=3$), $y_3 = D \frac{3\lambda/(4\lambda)}{\sqrt{1-(3\lambda/4\lambda)^2}} = D \frac{3/4}{\sqrt{1-9/16}} = D \frac{3/4}{\sqrt{7/16}} = D \frac{3/4}{\sqrt{7}/4} = \frac{3D}{\sqrt{7}}$. Option C is incorrect.

For the $1^{st}$ maxima ($n=1$), $y_1 = D \frac{1\lambda/(4\lambda)}{\sqrt{1-(1\lambda/4\lambda)^2}} = D \frac{1/4}{\sqrt{1-1/16}} = D \frac{1/4}{\sqrt{15/16}} = D \frac{1/4}{\sqrt{15}/4} = \frac{D}{\sqrt{15}}$. For the $2^{nd}$ maxima ($n=2$), $y_2 = D \frac{2\lambda/(4\lambda)}{\sqrt{1-(2\lambda/4\lambda)^2}} = D \frac{1/2}{\sqrt{1-1/4}} = D \frac{1/2}{\sqrt{3/4}} = D \frac{1/2}{\sqrt{3}/2} = \frac{D}{\sqrt{3}}$. The distance between the first and second maxima is $y_2 - y_1 = \frac{D}{\sqrt{3}} - \frac{D}{\sqrt{15}} = D \left(\frac{\sqrt{5}-1}{\sqrt{15}}\right) = \frac{D}{\sqrt{15}}(\sqrt{5}-1)$. Option D is correct.