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Light of wavelength $6000 Å$ falls on a single slit of width $0.1 \mathrm{~mm}$. The second minima will be formed for the angle of diffraction of
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The correct answer is:
0.012 radian
Here, wavelength $\lambda=6000 Å=6000 \times 10^{-10} \mathrm{~m}$
Slit width $a=0.1 \mathrm{~mm}=0.1 \times 10^{-3} \mathrm{~m}$
In case of diffraction at single slit, the position of minima is given by
$a \sin \theta_n=n \lambda$, where $n=1,2,3, \ldots$.
second minimum, $n=2 \quad \therefore \quad a \sin \theta_2=2 \lambda$
For small angle $\sin \theta_2 \approx \theta_2$.
$\therefore \theta_2=\frac{2 \lambda}{a}$
Substituting the values, we get
$\therefore \theta_2=\frac{2 \times 6000 \times 10^{-10}}{0.1 \times 10^{-3}}=0.012$ radian
Slit width $a=0.1 \mathrm{~mm}=0.1 \times 10^{-3} \mathrm{~m}$
In case of diffraction at single slit, the position of minima is given by
$a \sin \theta_n=n \lambda$, where $n=1,2,3, \ldots$.
second minimum, $n=2 \quad \therefore \quad a \sin \theta_2=2 \lambda$
For small angle $\sin \theta_2 \approx \theta_2$.
$\therefore \theta_2=\frac{2 \lambda}{a}$
Substituting the values, we get
$\therefore \theta_2=\frac{2 \times 6000 \times 10^{-10}}{0.1 \times 10^{-3}}=0.012$ radian
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