MHT CET · Physics · Wave Optics
A single slit of width \(d\) is illuminated by violet light of wavelength 400 nm and the width of the diffraction pattern is measured as ' \(Y\) '. When half of the slit width is covered and illuminated by yellow light of wavelength 600 nm , the width of the diffraction pattern is
- A zero
- B \(\frac{\mathrm{Y}}{3}\)
- C 3 Y
- D 4 Y
Answer & Solution
Correct Answer
(C) 3 Y
Step-by-step Solution
Detailed explanation
Width of diffraction pattern,
\(\mathrm{W}=\frac{2 \lambda \mathrm{D}}{\mathrm{~d}}\)
When half the slit is covered, \(\mathrm{d}^{\prime}=\frac{\mathrm{d}}{2}\) and \(\lambda^{\prime}=600 \mathrm{~nm}\)
\(\begin{array}{ll}
\therefore & \mathrm{W}^{\prime}=\frac{2 \lambda^{\prime}}{\mathrm{d} / 2} \\
\therefore & \frac{\mathrm{~W}^{\prime}}{\mathrm{W}}=\frac{2 \lambda^{\prime}}{\lambda}=\frac{2 \times 600}{400} \\
\therefore & \frac{\mathrm{~W}^{\prime}}{\mathrm{W}}=3 \\
\therefore & \mathrm{~W}^{\prime}=3 \mathrm{~W}=3 \mathrm{Y}
\end{array}\)
...(given, \(\mathrm{W}=\mathrm{Y}\) )
\(\mathrm{W}=\frac{2 \lambda \mathrm{D}}{\mathrm{~d}}\)
When half the slit is covered, \(\mathrm{d}^{\prime}=\frac{\mathrm{d}}{2}\) and \(\lambda^{\prime}=600 \mathrm{~nm}\)
\(\begin{array}{ll}
\therefore & \mathrm{W}^{\prime}=\frac{2 \lambda^{\prime}}{\mathrm{d} / 2} \\
\therefore & \frac{\mathrm{~W}^{\prime}}{\mathrm{W}}=\frac{2 \lambda^{\prime}}{\lambda}=\frac{2 \times 600}{400} \\
\therefore & \frac{\mathrm{~W}^{\prime}}{\mathrm{W}}=3 \\
\therefore & \mathrm{~W}^{\prime}=3 \mathrm{~W}=3 \mathrm{Y}
\end{array}\)
...(given, \(\mathrm{W}=\mathrm{Y}\) )
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