A right angled prism of refractive index mu_1 is placed in a rectangular block of refractive index m

Q: Why is the angle of incidence at the slanted face exactly $45^\circ$?

The prism is a right-angled triangle. Since the ray enters horizontally and the left face is vertical, it travels horizontally inside. The slanted face makes a $45^\circ$ angle with the horizontal, so the normal to this face also makes a $45^\circ$ angle with the horizontal ray.

Q: Why does path 'g' mean $\mu_1 = \mu_2$?

If a ray passes across an interface without changing its direction (zero deviation), it means the speed of light is the same in both media. According to Snell's law, this requires their refractive indices to be equal.

Q: How do we know path 'f' corresponds to $\mu_2 > \mu_1$?

Path 'f' bends towards the normal. Light bends towards the normal only when it slows down, which happens when it enters an optically denser medium. Therefore, the second medium ($\mu_2$) must have a higher refractive index than the first ($\mu_1$).

Q: Why is there an upper limit on $\mu_1$ for path 'h'?

Path 'h' bends away from the normal, meaning $\mu_1 > \mu_2$. However, if $\mu_1$ is too large, the critical angle becomes smaller than $45^\circ$, and the ray would undergo Total Internal Reflection (path 'i') instead of refracting. Thus, $\mu_1$ must be less than $\sqrt{2}\mu_2$.

Q: What role does $\mu_3$ play in this problem?

The condition $\mu_2 > \mu_3$ ensures that the ray can successfully emerge from the rectangular block into the surrounding medium without undergoing Total Internal Reflection at the outer boundary.

Question

A right angled prism of refractive index $\mu_1$ is placed in a rectangular block of refractive index $\mu_2$, which is surrounded by a medium of refractive index $\mu_3$, as shown in the figure. A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between $\mu_1, \mu_2$ and $\mu_3$, it takes one of the four possible paths 'ef', 'eg', 'eh' or 'ei'. Match the paths in Column I with conditions of refractive indices in Column II and select the correct answer using the codes given below the lists. \begin{array}{ll} extbf{Column I} & extbf{Column II} \ ext{P. } e ightarrow f & ext{1. } \mu_1 > \sqrt{2}\mu_2 \ ext{Q. } e ightarrow g & ext{2. } \mu_2 > \mu_1 ext{ and } \mu_2 > \mu_3 \ ext{R. } e ightarrow h & ext{3. } \mu_1 = \mu_2 \ ext{S. } e ightarrow i & ext{4. } \mu_2 < \mu_1 < \sqrt{2}\mu_2 ext{ and } \mu_2 > \mu_3 \end{array} Options: (A) P-2, Q-3, R-1, S-4 (B) P-1, Q-2, R-4, S-3 (C) P-4, Q-1, R-2, S-3 (D) P-2, Q-3, R-4, S-1

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine a ray of light, let's call it 'e', embarking on a journey through a series of optical media. It first encounters a rectangular block with a refractive index $\mu_2$. Because it strikes the vertical face at a perfect $90^\circ$ angle (normal incidence), it marches straight through without bending. Inside this block lies a right-angled isosceles prism with a refractive index $\mu_1$. The ray enters the vertical face of this prism, again normally, and continues its straight path until it hits the slanted hypotenuse. Because the prism is an isosceles right triangle, simple geometry tells us that the angle of incidence $i$ at this slanted face is exactly $45^\circ$. Now, the real magic begins. Depending on the optical densities (refractive indices) of the prism ($\mu_1$) and the surrounding block ($\mu_2$), the ray has four possible destinies: 'f', 'g', 'h', or 'i'. Let's decode each one. ### The Extreme Case: Total Internal Reflection (Path 'i') Let's look at path 'i'. The ray doesn't even cross the boundary; it reflects entirely back into the prism. This phenomenon is known as **Total Internal Reflection (TIR)**. For TIR to occur, the light must be attempting to travel from a denser medium to a rarer medium, and the angle of incidence must exceed the critical angle $ heta_c$. Mathematically, this is expressed as: $$i > heta_c$$ Since we know $i = 45^\circ$, we can write: $$\sin 45^\circ > \sin heta_c$$ From Snell's Law, we know that $\sin heta_c = \frac{\mu_2}{\mu_1}$. Substituting this in, we get: $$\frac{1}{\sqrt{2}} > \frac{\mu_2}{\mu_1}$$ Rearranging this inequality gives us the condition for path 'i': $$\mu_1 > \sqrt{2}\mu_2$$ This perfectly matches condition **1**, so **S $ ightarrow$ 1**. ### The Straight Path: No Deviation (Path 'g') Now, consider path 'g'. The ray passes through the slanted interface as if it wasn't even there! It doesn't bend towards or away from the normal. In optics, a ray only passes undeviated at an oblique angle if the two media have the exact same optical density. Therefore, the refractive index of the prism must be identical to the refractive index of the block: $$\mu_1 = \mu_2$$ This matches condition **3**, so **Q $ ightarrow$ 3**. ### Bending Towards the Normal (Path 'f') What if the ray takes path 'f'? Notice how the ray bends *towards* the normal line. According to Snell's Law, a light ray bends towards the normal when it slows down—that is, when it enters an optically denser medium from a rarer one. This tells us that the rectangular block is denser than the prism: $$\mu_2 > \mu_1$$ Furthermore, the problem states that the ray eventually emerges from the rectangular block into the surrounding medium $\mu_3$. For it to successfully exit without undergoing TIR at the outer boundary, the block must be denser than the outside medium, meaning $\mu_2 > \mu_3$. This matches condition **2**, so **P $ ightarrow$ 2**. ### Bending Away From the Normal (Path 'h') Finally, let's examine path 'h'. Here, the ray bends *away* from the normal. This indicates that the light is speeding up, transitioning from a denser medium to a rarer medium. Therefore: $$\mu_1 > \mu_2$$ However, unlike path 'i', the ray actually manages to refract and cross the boundary. It does *not* undergo Total Internal Reflection. This means the angle of incidence is less than the critical angle: $$i < heta_c \implies \sin 45^\circ < \sin heta_c \implies \frac{1}{\sqrt{2}} < \frac{\mu_2}{\mu_1}$$ Rearranging this gives $\mu_1 < \sqrt{2}\mu_2$. Combining our inequalities, we find the precise window for $\mu_1$: $$\mu_2 < \mu_1 < \sqrt{2}\mu_2$$ Again, to ensure the ray emerges from the block, we must have $\mu_2 > \mu_3$. This matches condition **4**, so **R $ ightarrow$ 4**. ### The Final Synthesis By systematically applying Snell's Law and the conditions for Total Internal Reflection, we have successfully mapped every possible path to its corresponding refractive index condition: * **P $ ightarrow$ 2** * **Q $ ightarrow$ 3** * **R $ ightarrow$ 4** * **S $ ightarrow$ 1** Looking at the given options, this sequence corresponds exactly to **option (d)**. This problem is a beautiful exercise in visualizing how light interacts with boundaries and how mathematical inequalities govern physical realities!

Select Answer:

Visualized Solution

\text{Analyzing the Setup}

\text{Path 'i' - Total Internal Reflection}

\text{Path 'g' - Undeviated Ray}

\text{Path 'f' - Bending Towards Normal}

\text{Path 'h' - Bending Away From Normal}

\text{Final Matching}

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Extreme Case

The Straight Path

Bending Towards the Normal (Path 'f')

Bending Away From the Normal (Path 'h')

The Final Synthesis

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