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Chapter 18: Problem 51

Light wave enters from medium 1 to medium \(2 .\) Its velocity in second medium is double from first. For total internal reflection, the angle of incidence must be greater than (A) \(30^{\circ}\) (B) \(60^{\circ}\) (C) \(45^{\circ}\) (D) \(90^{\circ}\)

Short Answer

Expert verified
The minimum angle for total internal reflection is greater than \(30^{\circ}\).

Step by step solution

01

Relationship between index of refraction and velocity

Recall that the index of refraction (\(n\)) of a medium is related to the speed of light in that medium (\(v\)) by the following equation: \[ n = \frac{c}{v} \] where \(c\) is the speed of light in a vacuum, \(v\) is the speed of light in the medium, and \(n\) is the index of refraction of the medium. Since the velocity of light in medium 2 is double that of medium 1, we can write: \[ v_2 = 2v_1 \] Now we can find the relationship between the indices of refraction in medium 1 (\(n_1\)) and medium 2 (\(n_2\)).
02

Find the relationship between the indices of refraction

Using the equation for index of refraction from Step 1 for both media, we have: \[ n_1 = \frac{c}{v_1} \text{ and } n_2 = \frac{c}{v_2} \] Since \(v_2=2v_1\): \[ n_2 = \frac{c}{2v_1} \] Substituting the expression for \(n_1\) in terms of \(v_1\): \[ n_2 = \frac{n_1}{2} \] This equation indicates that the index of refraction of medium 2 is half that of medium 1.
03

Apply Snell's law and find the critical angle

Snell's law relates the angles of incidence (\(\theta_1\)) and refraction (\(\theta_2\)) with the indices of refraction of the two media: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \] For total internal reflection to occur, we need to find the critical angle (\(\theta_c\)). At the critical angle, the refracted angle \(\theta_2 = 90^{\circ}\) (i.e., the refracted light wave is refracted along the boundary of the two media). With this condition, Snell's law becomes: \[ n_1 \sin \theta_c = n_2 \cdot 1 = n_2\] Now, we have to solve for the critical angle \(\theta_c\) using the relationship between \(n_1\) and \(n_2\) we derived in Step 2.
04

Solve for the critical angle

Using the relationship we derived in Step 2, \(n_2 = \frac{n_1}{2}\), substitute it into the modified Snell's law equation from Step 3: \[n_1 \sin \theta_c = \frac{n_1}{2} \] Simplifying the equation, we get: \[ \sin \theta_c = \frac{1}{2} \] Now, find the angle \(\theta_c\): \[ \theta_c = \sin^{-1}\left(\frac{1}{2}\right) \] \[ \theta_c = 30^{\circ} \] So, the critical angle for total internal reflection is \(30^{\circ}\).
05

Find the minimum angle for total internal reflection

For total internal reflection to occur, the angle of incidence must be greater than the critical angle. Therefore, the minimum angle for total internal reflection is: \[ \theta_{TIR} > 30^{\circ} \] Comparing our answer with the given options, the correct option is: (A) \(30^{\circ}\)

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