Question 36: Waves 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

$$2 \beta$$

Waves 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: