Optics 6 Question 9
10. The resolving power of electron microscope is higher than that of an optical microscope because the wavelength of electrons is than the wavelength of visible light.
(1992, 1M)
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Solution:
- In Young’s double slit experiment, the condition of bright fringe and dark fringe are,
for bright fringes (maxima), path difference $=n \lambda$
$$ d \sin \theta=n \lambda $$
for dark fringes (minima),
path-difference $=(2 n-1) \frac{\lambda}{2}$
$$ d \sin \theta=(2 n-1) \frac{\lambda}{2} $$
For the given question, distance between slits
$$
(d)=0.320 mm $$
Wavelength of light used $(\lambda)=500 n-m$
Angular range for bright fringe $(\theta)=-30^{\circ}$ to $30^{\circ}$
Hence, for bright-fringe,
$$ \begin{aligned} n \lambda & =d \sin \theta \\ n & =\frac{d \sin \theta}{\lambda}=\frac{0.320 \times 10^{-3} \times \sin 30^{\circ}}{500 \times 10^{-9}} \\ n _{\max } & =320 \end{aligned} $$
$\therefore$ Total number of maxima between the two lines are
Here, $\quad \begin{array}{ll}n & =\left(n _{\max } \times 2\right)+1 \ n & n=(320 \times 2)+1 \ n & =641\end{array}$
