ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2025
  • Easy
  • Numerical

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is \(\mathrm{x} \times 10^{-3} \mathrm{~cm}\), where x -value is ________.

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(A) 15

Step-by-step Solution

Detailed explanation

Width of 20 maxima of double slit \(=\) width of central maxima of single slit
\(\begin{aligned}
& \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\
& \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\
& \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm}
\end{aligned}\)
Value of \(x\) is 15
Answer is 15
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app