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Chapter 16: Problem 2245

A ray of light passes from glass \((\mathrm{n}=1.5)\) to medium \((\mathrm{n}=1.60)\) The value of the critical angle of glass is (A) \(\sin ^{-1}(16 / 15)\) (B) \(\sin ^{-1} \sqrt{(16 / 15)}\) (C) \(\sin ^{-1}(1 / 2)\) (D) \(\sin ^{-1}(15 / 16)\)

Short Answer

Expert verified
The short answer based on the provided step-by-step solution is: \(\theta_c = \sin^{-1} \left(\frac{16}{15}\right)\)

Step by step solution

01

Applying Snell's Law

To begin, we'll need to use Snell's Law, which relates the angles of incidence and refraction to the refractive index of each medium. Snell's Law is given by: \(n_1 \sin \theta _1 = n_2 \sin \theta _2\) where \(n_1\) is the refractive index of medium 1 (glass, in this case), \(\theta_1\) is the angle of incidence, \(n_2\) is the refractive index of medium 2, and \(\theta_2\) is the angle of refraction.
02

Finding the critical angle

When a light ray travels from a denser medium to a rarer medium, the critical angle is the angle of incidence where the angle of refraction becomes 90 degrees. So we have: \(n_1 \sin \theta_c = n_2 \sin 90^{\circ}\) Since \(\sin 90^{\circ} = 1\), we can simplify the equation as: \(n_1 \sin \theta_c = n_2\) Now, rearrange the equation to find the critical angle, \(\theta_c\): \(\sin \theta_c = \frac{n_2}{n_1}\)
03

Substituting the given values

Now, substitute the given values of refractive indexes \(n_1 = 1.5\) and \(n_2 = 1.6\) into the equation: \(\sin \theta_c = \frac{1.6}{1.5}\) This simplifies to: \(\sin \theta_c = \frac{16}{15}\) Now, find the inverse sine of the value to get the critical angle: \(\theta_c = \sin^{-1} \left(\frac{16}{15}\right)\) According to this result, the correct answer is option (A).

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Most popular questions from this chapter

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