Q.The phase difference between the two superimposing waves that give rise to a bright spot in a Young's double-slit experiment is ($n$ is an integer) : (A) $2n\pi$ (B) $2n\pi + \dfrac{\pi}{4}$ (C) $2n\pi + \dfrac{\pi}{2}$ (D) $2n\pi + \pi$

The heart of Young’s double-slit experiment is the interference of two coherent light waves. A bright spot — a constructive interference maximum — occurs where the two waves reinforce each other. Reinforcement happens when the crest of one wave meets the crest of the other, or trough meets trough. In other words, the waves must be in phase at that point.

The phase difference between two waves is directly linked to the path difference they travel. If the path difference is an integer multiple of the wavelength $\lambda$, the waves arrive in phase. Since one full wavelength corresponds to a phase change of $2\pi$ radians, the condition for a bright spot is:

where $n = 0, 1, 2, \dots$ (an integer).

Now let’s walk through the reasoning step by step.

1. What does “bright spot” mean physically?

   A bright fringe is a point on the screen where the intensity is maximum. In wave terms, this happens when the electric fields of the two waves add constructively — their amplitudes add. For this, the waves must be in phase at that point. If they were out of phase, they would partially or fully cancel, giving a dimmer or dark fringe.

2. How does phase difference arise?

   The two slits act as coherent sources. Light from each slit travels a different distance to reach a given point on the screen. This difference in path length creates a phase difference:

3. What path difference gives a bright spot? For constructive interference, the path difference must be an integer multiple of the wavelength:

Why? Because if one wave has traveled exactly $n$ whole wavelengths more than the other, its crests still align with the other’s crests — they are back in step.

4. Convert path difference to phase difference. Substituting $\Delta x = n\lambda$ into the phase formula: $\Delta \phi = \frac{2\pi}{\lambda} \times n\lambda = 2n\pi$ …

The heart of this problem lies in understanding how a medium affects the interference pattern. In Young's double-slit experiment, constructive interference (bright fringes) occurs when the optical path difference between rays from the two slits equals an integer multiple of the wavelength.

When we immerse the apparatus in a medium of refractive index $\mu$, the wavelength of light changes from $\lambda_{\text{air}}$ to $\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu}$. This compression of wavelength means more fringes fit into the same physical space on the screen.

The position of the $n$-th maximum is given by the condition that the path difference equals $n\lambda$. Since the geometry (slit separation $d$, screen distance $D$) remains fixed, we can compare which maxima land at the same physical position on the screen in the two cases.

Step-by-step reasoning:

1. Position of maxima in air: For the $n_{\text{air}}$-th maximum in air, the path difference is:

The position on the screen is $y = \frac{n_{\text{air}} \lambda_{\text{air}} D}{d}$.

2. Position of maxima in the medium: For the $n_{\text{medium}}$-th maximum in the medium (refractive index $\mu = \frac{4}{3}$), the wavelength becomes $\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu}$. The path difference is: $\Delta x = n_{\text{medium}} \lambda_{\text{medium}} = n_{\text{medium}} \frac{\lambda_{\text{air}}}{\mu}$ …

In Young’s double-slit experiment, the fringe width $\beta$ is the distance between two consecutive bright (or dark) fringes on the screen. The standard formula is:

where $\lambda$ is the wavelength of light in the medium, $D$ is the distance from the slits to the screen, and $d$ is the separation between the slits.

The key insight here is that when you change the medium, only the wavelength changes — $D$ and $d$ are physical distances that remain the same. The wavelength in a medium of refractive index $\mu$ is related to the wavelength in vacuum $\lambda_0$ by:

So the entire change in fringe width comes from this one factor.

Let’s walk through it step by step.

1. Write the fringe width in air (or vacuum). In air, the wavelength is $\lambda_0$ (approximately the same as in vacuum for most exam problems). So:

2. Write the fringe width in the liquid. In the liquid, the wavelength becomes $\lambda' = \lambda_0 / \mu$. The slit separation $d$ and screen distance $D$ are unchanged. So:

3. Substitute the original $\beta$. From step 1, $\frac{\lambda_0 D}{d} = \beta$. Therefore: …

Part (a)

In Young's double-slit experiment the fringe at a point is fixed by the optical path difference $\Delta x$ between the two waves:

• bright (constructive): $\Delta x=n\lambda$;
• dark (destructive): $\Delta x=(2n-1)\dfrac{\lambda}{2}$, i.e. $\dfrac{\lambda}{2},\dfrac{3\lambda}{2},\dfrac{5\lambda}{2},\dots$ for $n=1,2,3,\dots$

Given $\Delta x=\dfrac{5\lambda}{2}$:

The half-integer sequence $\tfrac{\lambda}{2}$ (1st), $\tfrac{3\lambda}{2}$ (2nd), $\tfrac{5\lambda}{2}$ (3rd) confirms this is the third dark fringe. …