1. Key idea: phase difference with and without sheet

In YDSE, the phase difference at point $P$ is determined by the optical path difference.

If a transparent sheet of refractive index $\mu$ and thickness $t$ is placed before slit $S_2$, then the optical path of the ray from $S_2$ increases by

$$(\mu-1)t.$$

So the net path difference at point $P$ becomes

$$\Delta(P)=S_1P-S_2P-(\mu-1)t$$

if we define geometric path difference as $S_1P-S_2P$.

Hence phase difference is

$$\delta(P)=\frac{2\pi}{\lambda}\,\Delta(P).$$

Also, intensity depends on phase difference as

$$I(P)=I_{\max}\cos^2\frac{\delta(P)}{2}$$

(for equal-intensity slits).

So:

• Bright fringe: $\delta=2n\pi$ i.e. $\Delta=n\lambda$
• Dark fringe: $\delta=(2n+1)\pi$ i.e. $\Delta=(2n+1)\lambda/2$

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2. Given geometric path differences

We are given:

$$S_1P_0=S_2P_0 \Rightarrow S_1P_0-S_2P_0=0$$ $$S_1P_1-S_2P_1=\frac{\lambda}{4}$$ $$S_1P_2-S_2P_2=\frac{\lambda}{3}$$

Let us test each case.

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3. Case (A): No sheet

Here $(\mu-1)t=0$.

So:

$$\Delta(P_0)=0,\qquad \delta(P_0)=0$$

Thus (P) is true.

At $P_1$:

$$\Delta(P_1)=\frac{\lambda}{4} Rightarrow \delta(P_1)=\frac{2\pi}{\lambda}\cdot \frac{\lambda}{4}=\frac{\pi}{2}$$

Hence

$$I(P_1)=I_{\max}\cos^2\frac{\pi}{4}=\frac{I_{\max}}{2}.$$

At $P_0$:

$$I(P_0)=I_{\max}.$$

Therefore

$$I(P_0)>I(P_1)$$

So (S) is true.

Thus for (A):

$$(A)\to (P),(S)$$

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4. Case (B): $(\mu-1)t=\lambda/4$

Now

$$\Delta(P)= (S_1P-S_2P)-\frac{\lambda}{4}.$$

At $P_1$:

$$\Delta(P_1)=\frac{\lambda}{4}-\frac{\lambda}{4}=0$$

So

$$\delta(P_1)=0.$$

Thus (Q) is true.

Check whether $I(P_1)=0$: No, because $\delta(P_1)=0$ gives maximum intensity, not zero. So (R) is false.

Thus for (B):

$$(B)\to (Q)$$

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5. Case (C): $(\mu-1)t=\lambda/2$

Now

$$\Delta(P)= (S_1P-S_2P)-\frac{\lambda}{2}.$$

At $P_1$:

$$\Delta(P_1)=\frac{\lambda}{4}-\frac{\lambda}{2}=-\frac{\lambda}{4}$$

Magnitude gives phase difference

$$|\delta(P_1)|=\frac{2\pi}{\lambda}\cdot \frac{\lambda}{4}=\frac{\pi}{2}.$$

So

$$I(P_1)=I_{\max}\cos^2\frac{\pi}{4}=\frac{I_{\max}}{2}.$$

At $P_2$:

$$\Delta(P_2)=\frac{\lambda}{3}-\frac{\lambda}{2}=-\frac{\lambda}{6}$$

So

$$|\delta(P_2)|=\frac{2\pi}{\lambda}\cdot \frac{\lambda}{6}=\frac{\pi}{3}.$$

Hence

$$I(P_2)=I_{\max}\cos^2\frac{\pi}{6}=I_{\max}\cdot \frac{3}{4}.$$

Therefore

$$I(P_2)=\frac{3I_{\max}}{4}>\frac{I_{\max}}{2}=I(P_1).$$

So (T) is true.

Thus for (C):

$$(C)\to (T)$$

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6. Case (D): $(\mu-1)t=3\lambda/4$

Now

$$\Delta(P)= (S_1P-S_2P)-\frac{3\lambda}{4}.$$

At $P_1$:

$$\Delta(P_1)=\frac{\lambda}{4}-\frac{3\lambda}{4}=-\frac{\lambda}{2}$$

So

$$|\delta(P_1)|=\frac{2\pi}{\lambda}\cdot \frac{\lambda}{2}=\pi$$

Hence

$$I(P_1)=0.$$

So (R) is true.

At $P_0$:

$$\Delta(P_0)=0-\frac{3\lambda}{4}=-\frac{3\lambda}{4}$$

So

$$|\delta(P_0)|=\frac{2\pi}{\lambda}\cdot \frac{3\lambda}{4}=\frac{3\pi}{2}.$$

Thus

$$I(P_0)=I_{\max}\cos^2\frac{3\pi}{4}=I_{\max}\cdot \frac{1}{2}.$$

Therefore

$$I(P_0)=\frac{I_{\max}}{2}>0=I(P_1)$$

So (S) is true.

At $P_2$:

$$\Delta(P_2)=\frac{\lambda}{3}-\frac{3\lambda}{4}=-\frac{5\lambda}{12}$$

So

$$|\delta(P_2)|=\frac{2\pi}{\lambda}\cdot \frac{5\lambda}{12}=\frac{5\pi}{6}.$$

Hence

$$I(P_2)=I_{\max}\cos^2\frac{5\pi}{12}.$$

This is positive, so certainly

$$I(P_2)>0=I(P_1).$$

Thus (T) is true.

Therefore for (D):

$$(D)\to (R),(S),(T)$$

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7. Final matching

We obtained:

• $(A)\to (P),(S)$
• $(B)\to (Q)$
• $(C)\to (T)$
• $(D)\to (R),(S),(T)$

This exactly matches Option A.

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8. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.