Question 38: Wave Optics

Question

In a single slit diffraction experiment, for slit width '$\alpha$' the width of the central maxima is '$\beta$'. If we double the slit width then the corresponding width of the central maxima will be:

Accepted Answer

In a single slit diffraction experiment, the width of the central maxima is given by the diffraction pattern formula derived from the Huygens-Fresnel principle. The position of the first minima in the diffraction pattern is given by:
$ a \sin(	heta) = n \lambda $
where $a$ is the slit width, $	heta$ is the angle of the diffraction minimum, $n$ is the order of the minimum (for the first minimum, $n=1$), and $\lambda$ is the wavelength of the light used.
The angular width of the central maxima is approximately twice the angle to the first minimum:
$ \Delta 	heta = \frac{2 \lambda}{a} $
When projecting this onto a screen at a distance $D$, the width of the central maximum can be approximated by:
$ \beta = 2D 	an(	heta) \approx 2D \left(\frac{\lambda}{a}ight) $
If the slit width $a$ is doubled, say to $2a$, then the new width of the central maximum $\beta'$ becomes:
$ \beta' = 2D \left(\frac{\lambda}{2a}ight) = \frac{\beta}{2} $
Thus, if the slit width is doubled, the corresponding width of the central maxima will be:
Option C: $\frac{\beta}{2}$.

Answer

$4 \beta$

Answer

$\beta$

Answer

$\frac{\beta}{2}$

Answer

$2 \beta$

Wave Optics

In a single slit diffraction experiment, for slit width '\alpha' the width of the central maxima is '\beta'. If we double the slit width then the corresponding width of the central maxima will be:

Choose the correct answer: