Question

In a double-slit experiment, green light $(\lambda =5303\mathring{\text{A}})$ falls on a double slit, having a separation of $19.44\mu \text{m}$ and a width of $4.05\mu \text{m}$. The number of bright fringes between the first and the second diffraction minima is :

• A. $10$
• B. $05$
• C. $04$
• D. $09$

Answer 1

The correct option is B $05$



Here, $d=19.44\mu m=19.44\times {10}^{-6}\text{m}$
$a=4.05\mu m=4.05\times {10}^{-6}\text{m}$
$\lambda =5303\mathring{\text{A}}=5.303\times {10}^{-7}\text{m}$

Let $P$ and $Q$ are two points where first and second diffraction minima are formed, respectively.

Now, the distances of points $P$ and $Q$, from the centre of the screen are,

$y_1=\frac{\lambda D}{a}=\frac{5.303\times {10}^{-7}D}{4.05\times {10}^{-6}}=0.131D$

And, $y_2=\frac{\lambda D}{a}=\frac{2\times 5.303\times {10}^{-7}D}{4.05\times {10}^{-6}}=0.262D$

For interference,

path difference at $P$ is,
$(\Delta y{)}_P=\frac{y_{1}d}{D}=\frac{0.131D\times 19.44\times {10}^{-6}}{D}$

$=4.8\lambda$

And, path difference at $P$ is,
$(\Delta y{)}_P=\frac{y_{1}d}{D}=\frac{0.262D\times 19.44\times {10}^{-6}}{D}$

$=9.6\lambda$

Now, path difference at $P$ is $4.8\lambda <5\lambda$ and path difference at $Q$ is $9.6\lambda <10\lambda$.

So, between points $P$ and $Q$, five bright fringes will be seen; they will be: $5^{\text{th}}\left(5\lambda \right),6^{\text{th}}\left(6\lambda \right),7^{\text{th}}\left(7\lambda \right),8^{\text{th}}\left(8\lambda \right),\text{and}{9}^{\text{th}}\left(9\lambda \right)$

Hence, $\left(B\right)$ is the correct answer.