Q136 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
The wavelength of a monochromatic light which is used in single slit diffraction is 800 nm . The width of the single slit for which the first minimum appears at $\theta=45^{\circ}$ on the screen will be:
- A$1.3 \mu \mathrm{~m}$
- Correct answerB$1.13 \mu \mathrm{~m}$
- C$2.13 \mu \mathrm{~m}$
- D$1.23 \mu \mathrm{~m}$
Solution
The condition for the first minimum in single slit diffraction is given by the formula $a \sin \theta = n \lambda$, where $a$ is the width of the slit, $\theta$ is the angle of diffraction, $n$ is the order of the minimum, and $\lambda$ is the wavelength of the light.
Given values are $\lambda = 800 \text{ nm} = 800 \times 10^{-9} \text{ m}$ and $\theta = 45^{\circ}$. For the first minimum, $n = 1$.
Substituting these values into the formula:
$a \sin 45^{\circ} = 1 \times 800 \times 10^{-9} \text{ m}$
$a \times \dfrac{1}{\sqrt{2}} = 800 \times 10^{-9} \text{ m}$
$a = 800 \times \sqrt{2} \times 10^{-9} \text{ m}$
Using $\sqrt{2} \approx 1.414$:
$a = 800 \times 1.414 \times 10^{-9} \text{ m}$
$a = 1131.2 \times 10^{-9} \text{ m}$
$a = 1.1312 \times 10^{-6} \text{ m} = 1.13 \mu \text{m}$
Answer: $1.13 \mu \mathrm{~m}$
Given values are $\lambda = 800 \text{ nm} = 800 \times 10^{-9} \text{ m}$ and $\theta = 45^{\circ}$. For the first minimum, $n = 1$.
Substituting these values into the formula:
$a \sin 45^{\circ} = 1 \times 800 \times 10^{-9} \text{ m}$
$a \times \dfrac{1}{\sqrt{2}} = 800 \times 10^{-9} \text{ m}$
$a = 800 \times \sqrt{2} \times 10^{-9} \text{ m}$
Using $\sqrt{2} \approx 1.414$:
$a = 800 \times 1.414 \times 10^{-9} \text{ m}$
$a = 1131.2 \times 10^{-9} \text{ m}$
$a = 1.1312 \times 10^{-6} \text{ m} = 1.13 \mu \text{m}$
Answer: $1.13 \mu \mathrm{~m}$
Hence, the correct answer is option: 'B'.