NEET 2022 Physics Refraction MCQ Question

Chapter: Ray Optics and Optical Instruments

Class: 12 Physics Topic: Critical Angle Difficulty: 🟡 Medium

✅ Ans: D — $\sin^{-1}(0.750)$

Given:

$$v_A=1.5\times10^8\,m/s$$ $$v_B=2.0\times10^8\,m/s$$

Step 1: Identify denser medium

We know:

$$n=\frac{c}{v}$$

So, lower speed → higher refractive index → denser medium.

Therefore:

$$A=\text{denser},\qquad B=\text{rarer}$$

Step 2: Apply critical-angle formula

For light travelling from denser to rarer medium:

$$\sin C=\frac{n_B}{n_A}$$

Using $n=c/v$:

$$\sin C =\frac{c/v_B}{c/v_A} =\frac{v_A}{v_B}$$ $$\sin C=\frac{1.5\times10^8}{2.0\times10^8} =0.75$$

Hence,

$$\boxed{C=\sin^{-1}(0.750)}$$

❌ Why other options are wrong?

• A: $\tan^{-1}(0.500)$ ❌ Critical angle uses sine relation.
• B: $\tan^{-1}(0.750)$ ❌ Wrong trigonometric relation.
• C: $\sin^{-1}(0.500)$ ❌ Correct speed ratio is $1.5/2=0.75$.
• D: $\sin^{-1}(0.750)$ ✅ Correct.

📌 NCERT Concept

$$\boxed{\sin C=\frac{n_{\text{rarer}}}{n_{\text{denser}}}}$$

Since $n=c/v$:

$$\boxed{\sin C=\frac{v_{\text{denser}}}{v_{\text{rarer}}}}$$

🧠 NEET Trick

Critical angle = Denser → Rarer

$$\boxed{\sin C=\frac{\text{speed in denser}}{\text{speed in rarer}}}$$ $$\boxed{\sin C=\frac{1.5}{2}=0.75}$$

Therefore,

$$\boxed{C=\sin^{-1}(0.750)}$$