JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2022
- Hard
- Numerical
In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by \(5 \times 10^{-2} \,m\) towards the slits, the change in fringe width is \(3 \times 10^{-3} \,cm\). If the distance between the slits is \(1 \,mm\), then the wavelength of the light will be_______ \(nm\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(600\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{\lambda D}{d}\)
\(\Delta \beta=\frac{\lambda}{ d } \Delta D\)
\(\lambda=\frac{\Delta \beta \cdot d}{\Delta D }\)
\(=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}}\)
\(=60 \times 10^{-8}=600 \times 10^{-9} \,m\)
\(=600 \,nm\)
\(\Delta \beta=\frac{\lambda}{ d } \Delta D\)
\(\lambda=\frac{\Delta \beta \cdot d}{\Delta D }\)
\(=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}}\)
\(=60 \times 10^{-8}=600 \times 10^{-9} \,m\)
\(=600 \,nm\)
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