A student uses two prisms, A and B, made of different materials but having the same prism angle (…
A student uses two prisms, A and B, made of different materials but having the same prism angle ($60^\circ$).
| Prism | Minimum Deviation |
|---|---|
| A | $34^\circ$ |
| B | $47^\circ$ |
(i)
Calculate the refractive index of the material of prism A.
(ii)
‘Since prism B has larger minimum deviation, the student claims it must also have the greater dispersive power.’ Is this statement always correct? Justify your answer.
Answer
Answer (i)
AIWritten by AI (gemini) - it can contain mistakes.
$n = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{60^\circ+34^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(47^\circ)}{\sin(30^\circ)} = \frac{0\cdot7314}{0\cdot5} = 1\cdot463 \approx 1\cdot47$.
Final answer: 1.47
Answer (ii)
AIWritten by AI (gemini) - it can contain mistakes.
The student's statement is not always correct. A larger minimum deviation indicates a higher mean refractive index $\mu$, whereas dispersive power $\omega = \frac{\mu_V - \mu_R}{\mu - 1}$ depends on the rate of variation of refractive index with wavelength (the difference $\mu_V - \mu_R$), which depends on the material of the prism.
From ISC 2027 Specimen Physics Paper 1, question 14.