A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of

Q: Why does the wavelength change when the apparatus is immersed in a liquid?

The speed of light decreases when it enters a denser medium, but its frequency remains constant (as it depends only on the source). Since $v = f \lambda$, a decrease in speed results in a proportional decrease in wavelength.

Q: Why do we use relative refractive index for the glass slab's path difference?

The glass slab replaces a thickness $t$ of the liquid, not air. Therefore, the extra optical path introduced is $(\mu_g - \mu_l)t$. When we factor out $\mu_l$, it becomes $\mu_l (\mu_g/\mu_l - 1)t$. The path difference in terms of the liquid medium is just $(\mu_g/\mu_l - 1)t$.

Q: Why did we equate the path difference to $\lambda'/2$ for the minimum?

The central point $O$ originally had zero path difference (a maximum). To make it a minimum, the new path difference introduced by the slab must be an odd multiple of $\lambda'/2$. The smallest thickness corresponds to the first minimum, which is exactly $\lambda'/2$.

Q: Why is my answer $1.579 \mu\text{m}$ while the exact answer is $1.575 \mu\text{m}$?

This is a classic rounding error! If you calculate $\lambda' \approx 4737 \text{ \AA}$ and the relative refractive index term as $0.15$, you get $1.579 \mu\text{m}$. However, if you keep the exact fractions ($\lambda/1.33$ and $0.20/1.33$), the $1.33$ terms cancel out perfectly, yielding the exact answer of $1.575 \mu\text{m}$.

Q: Does the fringe width change when the glass slab is introduced?

No, the fringe width $\omega = \lambda' D / d$ depends only on the wavelength in the surrounding medium, the slit separation, and the screen distance. The glass slab only shifts the entire fringe pattern without altering the distance between consecutive fringes.

Question

A double slit apparatus is immersed in a liquid of refractive index $1.33$. It has slit separation of $1	ext{ mm}$ and distance between the plane of slits and screen is $1.33	ext{ m}$. The slits are illuminated by a parallel beam of light whose wavelength in air is $6300	ext{ \AA}$.
(a) Calculate the fringe width.
(b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index $1.53$. Find the smallest thickness of the sheet to bring the adjacent minimum on the axis.

Sigma Percentile · Accepted Answer

The beauty of physics often lies in the elegant cancellation of terms, revealing a deeper symmetry in the problem. This classic JEE Advanced question from 1996 is a perfect example of how avoiding premature rounding can save you from a subtle trap!

### Analyzing the Setup
Imagine a standard Young's Double Slit Experiment (YDSE). Now, take the entire apparatus and submerge it in a liquid with a refractive index of $\mu_l = 1.33$.

We are given:
*   Slit separation, $d = 1 	ext{ mm} = 10^{-3} 	ext{ m}$
*   Distance to screen, $D = 1.33 	ext{ m}$
*   Wavelength in air, $\lambda = 6300 	ext{ \AA}$

When light enters a denser medium, its speed decreases, and consequently, its wavelength shrinks. The new wavelength in the liquid is:
$$ \lambda' = \frac{\lambda}{\mu_l} = \frac{6300}{1.33} 	ext{ \AA} \approx 4737 	ext{ \AA} $$

### The Master Equation for Fringe Width
The fringe width $\omega$ is the distance between consecutive bright or dark fringes. The formula remains the same, but we must use the new wavelength $\lambda'$:
$$ \omega = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu_l d} $$

Notice the brilliant design of the problem! The distance $D = 1.33 	ext{ m}$ and the refractive index $\mu_l = 1.33$ are numerically identical. They cancel out perfectly:
$$ \omega = \frac{6300 	imes 10^{-10} 	imes 1.33}{1.33 	imes 10^{-3}} = 6.3 	imes 10^{-4} 	ext{ m} = \mathbf{0.63 	ext{ mm}} $$

### The Glass Slab and the Rounding Trap
Next, a glass slab of thickness $t$ and refractive index $\mu_g = 1.53$ is placed in front of one of the slits. This introduces an additional optical path.

The path difference created by the slab *in the liquid medium* is:
$$ \Delta x = \left(\frac{\mu_g}{\mu_l} - 1ight) t = \left(\frac{1.53 - 1.33}{1.33}ight) t = \frac{0.20}{1.33} t $$

We want the adjacent minimum to shift to the central axis $O$. The central axis originally had zero path difference (a maximum). To become the first minimum, the new path difference must be exactly half a wavelength:
$$ \Delta x = \frac{\lambda'}{2} = \frac{\lambda}{2 \mu_l} $$

Equating the two expressions:
$$ \frac{0.20}{1.33} t = \frac{6300 	ext{ \AA}}{2 	imes 1.33} $$

### Final Calculation: The Exact Answer
Look at the equation above. The $1.33$ in the denominators cancels out beautifully! 
$$ 0.20 t = \frac{6300 	ext{ \AA}}{2} = 3150 	ext{ \AA} $$
$$ t = \frac{3150}{0.20} 	ext{ \AA} = 15750 	ext{ \AA} = \mathbf{1.575 \mu	ext{m}} $$

**The Trap:** If you had calculated $\lambda' \approx 4737 	ext{ \AA}$ and $\left(\frac{1.53}{1.33} - 1ight) \approx 0.15$, you would get $t = \frac{4737}{2 	imes 0.15} = 15790 	ext{ \AA} = 1.579 \mu	ext{m}$. This is a classic rounding error! By keeping the exact fractions, we bypassed the error and arrived at the pristine, exact answer of $1.575 \mu	ext{m}$.

Visualized Solution

$Analyzing the Setup$

$Wavelength in Liquid$

$Fringe Width$

$Path Difference due to Glass Slab$

$Condition for Minimum at Center$

$Exact Calculation of Thickness$

$The Way Forward$

The Sigma Insight: Interference of Waves

Analyzing the Setup Imagine a standard Young's Double Slit Experiment (YDSE)

The Master Equation for Fringe Width The fringe width $ω$ is the distance between consecutive bright or dark fringes

The Glass Slab and the Rounding Trap Next, a glass slab of thickness $t$ and refractive index $μ_{g} = 1.53$ is placed in front of one of the slits

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

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

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

.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; } Condition for Minimum at Center

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Exact Calculation of Thickness

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward