CBSE Class 10 Science 2023: question 39

Hints

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Step-by-step solution

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1. (i) Use $n=\frac{c}{v}$, so $v=\frac{c}{n}$.

   ½ mark

2. $v=\frac{3\times 10^8\text{m/s}}{2.42}\approx 1.24\times 10^8$ m/s

   ½ mark

3. (ii) For the same angle of incidence, a higher refractive index bends light more towards the normal, so the angle of refraction is smaller. Increasing order: $r_{\text{carbon disulphide}}<r_{\text{glass}}<r_{\text{water}}$.

   1 mark

4. (iii)(A)(a) Glass is optically denser.

   ½ mark

5. Light travels slower in glass ($2\times 10^8$ m/s) than in water ($2.25\times 10^8$ m/s), so glass has the higher refractive index.

   ½ mark

6. (iii)(A)(b) The ray goes straight into the glass without bending. For normal incidence $\angle i=0^{\circ }$, so $\angle r=0^{\circ }$ and the ray continues along the normal (only its speed decreases).

   1 mark

7. (iii)(B)(i) $n_{glass}=\frac{c}{v_{glass}}\Rightarrow c=n_{glass}{v}_{glass}=\frac{3}{2}\times 2\times 10^8=3\times 10^8$ m/s

   1 mark

8. (iii)(B)(ii) $v_{water}=\frac{c}{n_{water}}=\frac{3\times 10^8}{4/3}=\frac{9}{4}\times 10^8=2.25\times 10^8$ m/s

   1 mark

Answer: (i) ≈1.24 × 10⁸ m/s; (ii) r(CS₂) < r(glass) < r(water); (iii)(A) (a) glass (light is slower, higher n); (b) the ray passes undeviated since i = 0 gives r = 0; (iii)(B) vacuum: 3 × 10⁸ m/s; water: 2.25 × 10⁸ m/s

Revise the chapter

• Chapter: Light – Reflection and Refraction (notes, formulas, where marks are lost and every past question)
• Topic: Refraction of Light and Refractive Index