.2 \mathrm{~mm}$ is. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,KCET,NCERT,NEET,Wave Optics,Single Slit Diffraction, JEE Main Physics,Class 11 Physics,Class 12 Physics.">
.2 \mathrm{~mm}$ is. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,KCET,NCERT,NEET,Wave Optics,Single Slit Diffraction, JEE Main Physics,Class 11 Physics,Class 12 Physics.">
.2 \mathrm{~mm}$ is. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,KCET,NCERT,NEET,Wave Optics,Single Slit Diffraction, JEE Main Physics,Class 11 Physics,Class 12 Physics.">
The difraction pattern of a light of wavelength $400 \mathrm{~nm}$ difracting from a slit of width $0.2 \mathrm{~mm}$ is focused on the focal plane of a convex lens of focal length $100 \mathrm{~cm}$. The width of the $1^{\text {st }}$ secondary maxima will be :
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Step-by-Step Solution
Step 1: List the Known Quantities
• Wavelength of light, $ \lambda = 400 \times 10^{-9} \text{ m}$
• Slit width, $ a = 0.2 \times 10^{-3} \text{ m}$
• Focal length of the convex lens (effectively the screen distance), $D = 1 \text{ m}$
Step 2: Use the Formula for the Width of the First Secondary Maximum
For a single-slit diffraction pattern on a screen (or lens focal plane) at distance $D$, the approximate width of the first secondary maximum can be taken as:
$ \text{Width} = \frac{\lambda}{a} \times D $
Step 3: Substitute the Values
$ \text{Width} = \frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 1 \text{ m} $
Step 4: Calculate Step by Step
Compute the ratio
$ \frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} $:
$
\frac{400 \times 10^{-9}}{0.2 \times 10^{-3}}
= \frac{400}{0.2} \times \frac{10^{-9}}{10^{-3}}
= 2000 \times 10^{-6}
= 2 \times 10^{-3} \text{ m}
$
Multiply by $ D = 1 \text{ m}$:
$
2 \times 10^{-3} \text{ m} \times 1 \text{ m}^{-1}
= 2 \times 10^{-3} \text{ m}
= 2 \text{ mm}
$
Step 5: Final Result
The width of the first secondary maximum in the diffraction pattern is
$2 \text{ mm}.$