Q.41. One side of an equilateral prism is painted by a transparent material of refractive index . Th

Q: What is the correct answer for this exam Physics Q41?

The correct answer is D. For total internal reflection (TIR) to occur at the painted face, the angle of incidence inside the prism must be greater than the critical angle ($i > i_c$). The critical angle is given by $\sin i_c = \frac{n_2}{n_1}$, where $n_1$ is the refractive index of the denser medium (prism) and $n_2$ is the refractive index of the rarer medium (painted material). Given: $n_{\text{prism}} = 1.6$. The pris

Q: Q.41. One side of an equilateral prism is painted by a transparent material of refractive index . The refractive index of prism is . The minimum value of required for total internal reflection from…

The correct answer is D. For total internal reflection (TIR) to occur at the painted face, the angle of incidence inside the prism must be greater than the critical angle ($i > i_c$). The critical angle is given by $\sin i_c = \frac{n_2}{n_1}$, where $n_1$ is the refractive index of the denser medium (prism) and $n_2$ is the refractive index of the rarer medium (painted material). Given: $n_{\text{prism}} = 1.6$. The pris

Q: What are the options for this this exam Physics question?

A. $3\sqrt{3}/1.6$ · B. $\sqrt{3}$ · C. $3.2/\sqrt{3}$ · D. $4\sqrt{3}/5$

Question

Q.41. One side of an equilateral prism is painted by a transparent material of refractive index . The refractive index of prism is . The minimum value of  required for total internal reflection from painted face is __________

FirstInTest · Accepted Answer

For total internal reflection (TIR) to occur at the painted face, the angle of incidence inside the prism must be greater than the critical angle ($i > i_c$). The critical angle is given by $\sin i_c = \frac{n_2}{n_1}$, where $n_1$ is the refractive index of the denser medium (prism) and $n_2$ is the refractive index of the rarer medium (painted material). Given: $n_{ ext{prism}} = 1.6$. The prism is equilateral, so all angles are $60^\circ$. Consider a ray entering the prism normally from one face. It will pass undeviated and strike the opposite face at an angle of incidence of $60^\circ$ (since the normal to the face makes an angle of $60^\circ$ with the incident ray, or by geometry, the angle of incidence at the second face is $60^\circ$). For TIR, $i > i_c$, so $\sin i > \sin i_c$. $\sin 60^\circ > \frac{n_2}{n_{ ext{prism}}}$. $\frac{\sqrt{3}}{2} > \frac{n_2}{1.6}$. $n_2 < 1.6 imes \frac{\sqrt{3}}{2} = 0.8 imes \sqrt{3} \approx 0.8 imes 1.732 = 1.3856$. This calculation is for a ray entering normally. The question asks for the minimum value of $n_2$ required for TIR. This means we need the maximum possible critical angle, which implies the minimum $n_2$. The condition for TIR is $n_{ ext{prism}} \sin i_c = n_2 \sin 90^\circ$, so $n_2 = n_{ ext{prism}} \sin i_c$. For TIR to occur, $i_c$ must be less than or equal to the angle of incidence. The maximum angle of incidence inside the prism at the painted face is $60^\circ$ (if light enters normally from another face). So, $i_c \le 60^\circ$. This means $\sin i_c \le \sin 60^\circ = \frac{\sqrt{3}}{2}$. Therefore, $n_2 = n_{ ext{prism}} \sin i_c \le 1.6 imes \frac{\sqrt{3}}{2} = 0.8 imes \sqrt{3} = 1.3856$. The minimum value of $n_2$ for which TIR *can* occur (i.e., $n_2$ must be less than $1.3856$) is not what the question asks. It asks for the minimum value of $n_2$ *required* for TIR. This implies that for any ray incident on the painted face, TIR should occur. This is usually interpreted as the critical angle being less than or equal to the minimum possible angle of incidence at that face. However, the standard interpretation for such problems is that for a ray incident at $60^\circ$ (as shown in the diagram, or implied by the equilateral prism), TIR should occur. So, $n_2$ must be less than $1.3856$. The options are all greater than 1. This suggests that the question might be asking for the condition where TIR *just* occurs for a specific angle, or it's phrased in a way that implies $n_2$ is the refractive index of the *surrounding medium* for TIR to happen from the prism to the surrounding medium. If $n_2$ is the refractive index of the painted material, it must be less than $n_{ ext{prism}}$. Let's re-examine the diagram. It shows a ray incident on the painted face at $60^\circ$. For TIR, $n_{ ext{prism}} \sin 60^\circ > n_2 \sin 90^\circ$. $1.6 imes \frac{\sqrt{3}}{2} > n_2$. $n_2 < 0.8 imes \sqrt{3} \approx 1.3856$. This means $n_2$ must be less than $1.3856$. None of the options satisfy this if we are looking for $n_2$ to be less than this value. The options are $3\sqrt{3}/1.6 \approx 3 imes 1.732 / 1.6 = 5.196 / 1.6 = 3.2475$, $\sqrt{3} \approx 1.732$, $3.2/\sqrt{3} \approx 3.2 / 1.732 = 1.847$, $4\sqrt{3}/5 \approx 4 imes 1.732 / 5 = 6.928 / 5 = 1.3856$. The last option $4\sqrt{3}/5$ is exactly $1.3856$. If the question asks for the *minimum value of $n_2$ required for total internal reflection*, it means that for $n_2$ to be this value, TIR *can* occur. If $n_2$ is exactly $1.3856$, then the critical angle is $60^\circ$. If $n_2$ is slightly less than $1.3856$, TIR will occur. If $n_2$ is slightly more than $1.3856$, TIR will not occur for $60^\circ$. So, the maximum value of $n_2$ for which TIR can occur at $60^\circ$ is $1.3856$. The question is phrased as 'minimum value of $n_2$ required'. This is confusing. If $n_2$ is very small, TIR is easier. If $n_2$ is large, TIR is harder. So, if we want TIR to happen, $n_2$ must be below a certain threshold. The threshold is $1.6 \sin 60^\circ = 1.3856$. So $n_2$ must be $\le 1.3856$. The maximum possible value for $n_2$ for TIR to occur is $1.3856$. If the question means 'what is the maximum $n_2$ for which TIR is possible at $60^\circ$', then $4\sqrt{3}/5$ is the answer. Given the options, this is the most plausible interpretation.

Answer

$3\sqrt{3}/1.6$

Answer

$\sqrt{3}$

Answer

$3.2/\sqrt{3}$

Answer

$4\sqrt{3}/5$

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