Question

Let's say when beam of light is passing through reactangular slit of width $a$ then after diffraction its width becomes $11a$ after travelling distance $d$, then distance $d$ for monochromatic light of wavelength $\lambda$ is $N$km, then $N$ is ($a = 1$ mm, $\lambda = 2.5$ nm) (Consider width of central maxima as sum of spread due to diffraction and slit width)

• A. 2.2
• B. 1.1
• C. 3.3
• D. 4.4

Answer 1

Answer: (A)

2.2

Answer 2

The width of the central maximum ($W$) in a single-slit diffraction pattern is given by the formula $W = \frac{2\lambda d}{a}$. The problem states that the diffracted beam width is $11a$, so $W = 11a$. Equating the two expressions: $11a = \frac{2\lambda d}{a}$. Solving for $d$: $d = \frac{11a^2}{2\lambda}$. Given $a = 1$ mm $= 1 \times 10^{-3}$ m and $\lambda = 2.5$ nm $= 2.5 \times 10^{-9}$ m. Substituting these values: $d = \frac{11 \times (1 \times 10^{-3} \text{ m})^2}{2 \times (2.5 \times 10^{-9} \text{ m})} = \frac{11 \times 10^{-6} \text{ m}^2}{5 \times 10^{-9} \text{ m}} = 2.2 \times 10^3$ m $= 2200$ m. Since $d = N$ km, $N = \frac{2200}{1000} = 2.2$.