Question

The plot of deviation ($\delta$) vs angle of incidence (i) for a prism, made of glass of refractive index $\mu = 2\cos(16^\circ)$, kept in air, is shown below. Find the minimum angle of deviation suffered by a light ray after passing through the prism in degrees.

Answer 1

20

Answer 2

The minimum angle of deviation ($\delta_{min}$) for a prism is related to the angle of incidence ($i$) and the prism angle ($A$) by the formula: $\delta_{min} = 2i_{min} - A$, where $i_{min}$ is the angle of incidence at minimum deviation.

From the given plot, we observe that the deviation is $28^\circ$ for two angles of incidence: $18^\circ$ and $42^\circ$. The angle of incidence for minimum deviation ($i_{min}$) is the average of these two angles: $i_{min} = \frac{18^\circ + 42^\circ}{2} = \frac{60^\circ}{2} = 30^\circ$.

Now, we use the refractive index formula at minimum deviation: $\mu = \frac{\sin\left(\frac{A + \delta_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Since at minimum deviation, $i_{min} = \frac{A + \delta_{min}}{2}$, we have: $\mu = \frac{\sin(i_{min})}{\sin(A/2)}$

We are given $\mu = 2\cos(16^\circ)$ and we found $i_{min} = 30^\circ$. Substituting these values: $2\cos(16^\circ) = \frac{\sin(30^\circ)}{\sin(A/2)}$ $2\cos(16^\circ) = \frac{1/2}{\sin(A/2)}$ $\sin(A/2) = \frac{1}{4\cos(16^\circ)}$

Using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$, we can also write the minimum deviation formula as: $\delta_{min} = 2\arcsin(\mu \sin(A/2)) - A$

Let's try to find the angle of prism $A$ using the value of $\sin(A/2)$: $\sin(A/2) = \frac{1}{4\cos(16^\circ)} \approx \frac{1}{4 \times 0.9613} \approx \frac{1}{3.8452} \approx 0.2600$ $A/2 \approx \arcsin(0.2600) \approx 15.07^\circ$ $A \approx 30.14^\circ$

Now, we can calculate the minimum deviation: $\delta_{min} = 2i_{min} - A = 2(30^\circ) - 30.14^\circ = 60^\circ - 30.14^\circ = 29.86^\circ$

However, looking at the graph, the minimum deviation is clearly less than $28^\circ$. This suggests there might be a specific angle of prism intended that simplifies the calculation or a common assumption for prism problems. A common prism angle is $60^\circ$. Let's check if $A=60^\circ$ is consistent with the given $\mu$.

If $A=60^\circ$, then $A/2 = 30^\circ$. $\mu = \frac{\sin(i_{min})}{\sin(A/2)} \implies \mu = \frac{\sin(30^\circ)}{\sin(30^\circ)} = 1$. But $\mu = 2\cos(16^\circ) \approx 1.92$, which is not $1$. So $A \neq 60^\circ$.

Let's re-examine the relation $\sin(A/2) = \frac{1}{4\cos(16^\circ)}$. There might be a relationship between $A$ and $16^\circ$. Consider the case where $A = 2 \times 16^\circ = 32^\circ$. Then $A/2 = 16^\circ$. $\sin(16^\circ) = \frac{1}{4\cos(16^\circ)}$ $4\sin(16^\circ)\cos(16^\circ) = 1$ $2(2\sin(16^\circ)\cos(16^\circ)) = 1$ $2\sin(32^\circ) = 1$ $\sin(32^\circ) = 0.5$, which is false.

Let's reconsider the formula $\delta_{min} = 2i_{min} - A$. From the graph, the minimum deviation appears to be around $20^\circ$. If $\delta_{min} = 20^\circ$ and $i_{min} = 30^\circ$, then $A = 2i_{min} - \delta_{min} = 2(30^\circ) - 20^\circ = 60^\circ - 20^\circ = 40^\circ$. Let's check if $A=40^\circ$ is consistent with $\mu = 2\cos(16^\circ)$. If $A=40^\circ$, then $A/2 = 20^\circ$. $\mu = \frac{\sin(i_{min})}{\sin(A/2)} = \frac{\sin(30^\circ)}{\sin(20^\circ)} = \frac{0.5}{0.342} \approx 1.46$. This is not equal to $2\cos(16^\circ) \approx 1.92$.

Let's assume that the angle of prism is $A = 60^\circ$. This is a standard assumption in many prism problems if not explicitly stated otherwise, and it is often implied by the context or typical problem setups. Given the provided refractive index and the shape of the deviation curve, it's highly probable that $A=60^\circ$ is intended, even if the calculations with the given $\mu$ don't perfectly align with the visual minimum deviation from the plot.

If we assume $A=60^\circ$, then the minimum deviation is calculated as: $\delta_{min} = 2\arcsin(\mu \sin(A/2)) - A$ $\delta_{min} = 2\arcsin(2\cos(16^\circ) \sin(30^\circ)) - 60^\circ$ $\delta_{min} = 2\arcsin(\cos(16^\circ)) - 60^\circ$ $\delta_{min} = 2\arcsin(\sin(74^\circ)) - 60^\circ$ $\delta_{min} = 2(74^\circ) - 60^\circ = 148^\circ - 60^\circ = 88^\circ$. This result is inconsistent with the graph.

There seems to be an inconsistency in the problem statement or the provided graph. However, if we strictly interpret the graph and the given deviation points: $i_{min} = 30^\circ$. The minimum deviation $\delta_{min}$ is the lowest point on the curve. Visually, it appears to be around $20^\circ$. Let's assume $\delta_{min} = 20^\circ$.

The question asks for the minimum angle of deviation. Based on the visual representation of the plot, the lowest point of the deviation curve is approximately $20^\circ$. The given refractive index and the deviation points might lead to a complex calculation of $A$, and the visual estimation of $\delta_{min}$ from the graph is the most direct approach to answer the question as posed, assuming the graph is representative.