In Young's experiment, the source is red light of wavelength 7 x 10^-7 m. When a thin glass plate of

Q: Why does the central fringe shift when a glass slab is introduced?

The glass slab slows down the light passing through it, increasing its optical path length. To compensate and have zero phase difference, the central maximum shifts towards the slit covered by the slab.

Q: Why is the new shift equated to 6 *red* fringes even though we are using green light?

The problem states the central fringe shifts to a position *initially occupied* by the 6th bright fringe *due to red light*. This is a specific physical location on the screen, defined by the red light's interference pattern.

Q: Does the refractive index of glass change with the color of light?

Yes, this phenomenon is called dispersion. The refractive index is generally higher for shorter wavelengths (like green or blue) compared to longer wavelengths (like red).

Q: How did we find the initial fringe width without knowing $D$ and $d$?

We used the information that the shift of 5 red fringes corresponds to a physical distance of $10^{-3}\text{ m}$. Since the shift is exactly 5 fringe widths, we simply divided the distance by 5.

Q: Why is the change in fringe width negative?

Green light has a shorter wavelength than red light. Since fringe width is directly proportional to wavelength, the green fringes are narrower, resulting in a negative change.

Question

In Young's experiment, the source is red light of wavelength $7 	imes 10^{-7}	ext{ m}$. When a thin glass plate of refractive index $1.5$ at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by $10^{-3}	ext{ m}$ to the position previously occupied by the 5th bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength $5 	imes 10^{-7}	ext{ m}$, the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of glass for green light. Also estimate the change in fringe width due to the change in wavelength.

Sigma Percentile · Accepted Answer

This problem is a classic exploration of Young's Double-Slit Experiment (YDSE) and the profound effects of introducing a transparent medium into the path of one of the interfering light waves. It beautifully ties together the concepts of optical path length, interference fringe shifts, and the phenomenon of dispersion.

### The Optical Path Length and Fringe Shift

Imagine the two light waves racing towards the screen from slits $S_1$ and $S_2$. In a standard setup, the central maximum forms exactly at the center $O$ because both waves travel the same distance in the same time. However, when we place a glass slab of thickness $t$ and refractive index $\mu$ in front of one slit, the light slows down as it travels through the glass.

This delay creates an extra **optical path difference** between the two waves, given by the elegant formula:

$$\Delta x = (\mu - 1)t$$

Because of this extra path difference, the entire fringe pattern shifts on the screen. The problem states that the central bright fringe moves to the position previously occupied by the 5th bright fringe. This means the extra path difference introduced by the slab is exactly equal to five times the wavelength of the red light:

$$(\mu - 1)t = 5\lambda_{	ext{red}}$$

By substituting the given values ($\mu = 1.5$ and $\lambda_{	ext{red}} = 7 	imes 10^{-7}	ext{ m}$), we can easily solve for the thickness of the glass plate, yielding **$t = 7 	imes 10^{-6}	ext{ m}$**.

### The Magic of Dispersion

Next, the experiment takes a fascinating turn. The red light is swapped for green light. You might expect the refractive index of the glass to remain $1.5$, but nature is more complex! Glass behaves differently for different colors of light, a phenomenon known as **dispersion**.

Let's call the new refractive index for green light $\mu'$. The problem tells us that with green light, the new shift corresponds to the position initially occupied by the **6th bright fringe due to red light**. This is a crucial detail! The physical location on the screen is defined by the red light's interference pattern. Therefore, we equate the new path difference to six times the *red* wavelength:

$$(\mu' - 1)t = 6\lambda_{	ext{red}}$$

Substituting the thickness $t$ we found earlier, we can solve for the new refractive index, finding **$\mu' = 1.6$**. As expected, the refractive index is higher for the shorter wavelength (green) compared to the longer wavelength (red).

### Fringe Width Dynamics

Finally, we need to determine the change in the fringe width. We are given a vital piece of information: the initial shift of 5 red fringes corresponds to a physical distance of $10^{-3}	ext{ m}$. This allows us to calculate the width of a single red fringe:

$$5\omega_{	ext{red}} = 10^{-3}	ext{ m} \implies \omega_{	ext{red}} = 0.2 	imes 10^{-3}	ext{ m}$$

The formula for fringe width is $\omega = \frac{\lambda D}{d}$. Since the geometric setup ($D$ and $d$) remains constant, the fringe width is directly proportional to the wavelength ($\omega \propto \lambda$). We can use this proportionality to find the green fringe width:

$$\omega_{	ext{green}} = \omega_{	ext{red}} 	imes \frac{\lambda_{	ext{green}}}{\lambda_{	ext{red}}} \approx 0.143 	imes 10^{-3}	ext{ m}$$

The change in fringe width is simply the difference between the new and old widths:

$$\Delta \omega = \omega_{	ext{green}} - \omega_{	ext{red}} = -5.71 	imes 10^{-5}	ext{ m}$$

The negative sign perfectly aligns with our physical intuition: green light has a shorter wavelength than red light, so its interference fringes are packed more closely together!

Visualized Solution

$Path difference due to glass slab$

$Shift of central fringe$

$Calculating thickness t$

$Second case with green light$

$Calculating new refractive index μ^{'}$

$Initial fringe width$

$Fringe width proportionality$

$Calculating new fringe width$

$Change in fringe width$

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Optical Path Length and Fringe Shift

The Magic of Dispersion

Fringe Width Dynamics

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path difference due to glass slab

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Shift of central fringe

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating thickness t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second case with green light

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating new refractive index μ′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial fringe width

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fringe width proportionality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating new fringe width

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Change in fringe width