a) In Youngs double slits experiment, what changes will you observe in fringe width for the following operations : i) When experiment is shifted from air to inside water [1] ii) When slits width is…

a)

i) Fringe width when experiment is shifted from air to water

In Young’s double-slit experiment, the fringe width $\beta$ is given by: $$\beta =\frac{\lambda D}{d}$$ where:

• $\lambda$ = wavelength of light,
• $D$ = distance between slits and screen,
• $d$ = separation between slits.

When the experiment is shifted from air to water, the wavelength of light decreases because the refractive index of water ($n\approx 1.33$) is greater than that of air ($n\approx 1$). The relationship between wavelength in air (${\lambda }_{\text{air}}$) and in water (${\lambda }_{\text{water}}$) is: $${\lambda }_{\text{water}}=\frac{{\lambda }_{\text{air}}}{n}$$ Since $\lambda$ decreases, the fringe width $\beta$ also decreases (inversely proportional to $\lambda$).

ii) Fringe width when slit width is increased

The fringe width $\beta$ depends on the separation between slits ($d$), not the width of each slit. Increasing the slit width affects the intensity distribution but does not change the fringe width because $\beta$ is determined by $d$, $D$, and $\lambda$.

b)

i) Diffraction of light

Diffraction is the bending of light waves around the corners of an obstacle or aperture, causing them to spread out and form a pattern of bright and dark fringes on a screen. It occurs when light passes through a narrow slit or around an edge, demonstrating its wave nature.

ii) Width of central diffraction peak

Given:

• Slit width $a=0.01\text{m}$,
• Distance to screen $D=4\text{m}$,
• Wavelength $\lambda =600\text{nm}=600\times 10^{-9}\text{m}$.

The angular position of the first minimum (edge of central peak) is given by: $$\sin \theta =\frac{\lambda }{a}$$ For small angles, $\sin \theta \approx \theta$ (in radians), so: $$\theta \approx \frac{\lambda }{a}=\frac{600\times 10^{-9}}{0.01}=6\times 10^{-5}\text{rad}$$ The width of the central peak on the screen is: $$\text{Width}=2D\theta =2\times 4\times 6\times 10^{-5}=4.8\times 10^{-4}\text{m}=0.48\text{mm}$$

Final result: 0.48 mm

OR

a) Changing a spherical wavefront into a plane wavefront

A spherical wavefront can be converted into a plane wavefront by using a convex lens. The lens converges the diverging spherical wavefront to a focal point, and beyond the focal point, the wavefront approximates a plane wave.

b) Huygens’ Principle and Law of Reflection

Huygens’ Principle states that every point on a wavefront acts as a source of secondary wavelets. To verify the law of reflection, consider an incident ray striking a reflecting surface:

1. Draw the incident wavefront $AB$ striking the surface at point $O$.
2. Draw the reflected wavefront $CD$ such that secondary wavelets from $O$ propagate in all directions.
3. The reflected ray follows the direction of the secondary wavelet that makes equal angles with the normal.

flowchart TD
    A["Incident wavefront (AB)"] -->|"Strikes at O"| B["Reflecting surface"]
    B --> C["Secondary wavelets"]
    C --> D["Reflected wavefront (CD)"]
    E["Normal at O"] -->|"θ_i = θ_r"| F["Law of reflection"]

Steps:

1. Draw the incident ray and normal at the point of incidence.
2. Construct secondary wavelets from all points on the wavefront.
3. The reflected ray is tangent to the last wavelet, forming equal angles with the normal.

This confirms that the angle of incidence (${\theta }_i$) equals the angle of reflection (${\theta }_r$).