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Chapter 16: Problem 2254

Angle of minimum deviation for a prism refractive index \(1.5\) is equal to the angle of the prism. Then the angle of prism (given, $\sin 48^{\circ} 36^{\prime}=0.75$ ) (A) \(62^{\circ}\) (B) \(82^{\circ}\) (C) \(60^{\circ}\) (D) \(41^{\circ}\)

Short Answer

Expert verified
The angle of the prism is \(60^{\circ}\), which corresponds to answer choice (C).

Step by step solution

01

Write down the formula for angle of minimum deviation in a prism

For a prism with refractive index n, angle of prism A, and angle of minimum deviation \(\delta\), the formula is: \(n = \frac{\sin(A + \delta) / 2}{\sin{A / 2}}\) Here, we are given \(n = 1.5\) and \(\delta = A\).
02

Substitute the given values into the formula

Substitute the given values into the formula: \(1.5 = \frac{\sin(A + A) / 2}{\sin{A / 2}}\) Since \(\delta = A\), we have: \(1.5 = \frac{\sin(2A) / 2}{\sin{A / 2}}\)
03

Eliminate the denominators

Now, we can eliminate the denominators by multiplying both sides by \(2\sin(A / 2)\): \(3\sin(A / 2) = \sin(2A)\)
04

Write the sine of double angle in terms of sin and cos of the angle

Recall the double-angle formula for sine: \(\sin(2A) = 2\sin(A)\cos(A)\) Now substitute this back into our equation: \(3\sin(A / 2) = 2\sin(A)\cos(A)\)
05

Use given sine of angle of minimum deviation

We know that for the angle of minimum deviation, its sine is: \(\sin(48^\circ 36') = 0.75\) Since the angle of minimum deviation is equal to the prism angle, A, we have: \(\sin(A) = 0.75\)
06

Solve for angle A

Now, we need to rewrite our equation from step 4 using the given sine of angle A and solve for angle A: \(3\sin(A/2)=2(0.75)\cos(A)\) \(3\sin(A/2)=1.5\cos(A)\) Since we know that \(\sin^2(x) + \cos^2(x) = 1\), we can find the cosine of angle A: \(\cos(A)=\sqrt{1-\sin^2(A)}=\sqrt{1-0.75^2}\) Now substitute the cosine of angle A into the equation: \(3\sin(A/2)=1.5\sqrt{1-0.75^2}\) Solving for \(\sin(A/2)\), we find: \(\sin(A/2)=\frac{1.5\sqrt{1-0.75^2}}{3}=\frac{1}{2}\) Finally, we can find the angle A: \(A=2\arcsin(\frac{1}{2})\) \(A=2 * 30^{\circ}\) \(A=60^{\circ}\) So, the angle of the prism is \(\boxed{60^{\circ}}\), which corresponds to answer choice (C).

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Most popular questions from this chapter

A mark at the bottom of the liquid appears to rise by \(0.2 \mathrm{~m}\), If depth of the liquid is \(2.0 \mathrm{~m}\) then refractive index of the liquid is (A) \(1.80\) (B) \(1.60\) (C) \(1.33\) (D) \(1.11\)

A light of wavelength \(320 \mathrm{~nm}\) enters in a medium of refractive index \(1.6\) from the air of refractive index \(1.0\). The new wavelength of light in the medium will be \(\mathrm{nm}\). (A) 520 (B) 400 (C) 320 (D) 200

$$ \begin{array}{|l|l|} \hline \text { Column - I } & \text { Column - II } \\ \hline \text { (i) While going from rarer to denser medium } & \text { (a) Wavelength changes } \\ \text { (ii) While going from denser to rarer medium } & \text { (b) } \eta=(\mathrm{C} / \mathrm{V}) \\ \text { (iii) While going to one medium to another } & \text { (C) Ray bends towards normal } \\ \text { (iv) Refractive index of medium } & \text { (D) Rav bends awav from normal } \\ \hline \end{array} $$ (A) \(i-c\), ii \(-d\), iii \(-b\), iv-a (B) \(\mathrm{i}-\mathrm{a}\), ii \(-\mathrm{b}\), iii $-\mathrm{c}, \mathrm{iv}-\mathrm{d}$ (C) $\mathrm{i}-\mathrm{c}, \mathrm{ii}-\mathrm{b}, \mathrm{iii}-\mathrm{a}, \mathrm{iv}-\mathrm{d}$ (D) \(i-d, 1 i-c, 11 i-b, i v-a\)

A convex lens of glass \((\mathrm{n}=1.5)\) has focal length \(0.2 \mathrm{~m}\). The lens is immersed in water of refractive index \(1.33\). The change in the power of convex lens is (A) \(3.72 \mathrm{D}\) (B) \(4.62 \mathrm{D}\) (C) \(6.44 \mathrm{D}\) (D) \(1.86 \mathrm{D}\)

A ray of light is incident at an angle \(30^{\circ}\) on a mirror, The angle between normal and reflected ray is (A) \(15^{\circ}\) (B) \(30^{\circ}\) (C) \(45^{\circ}\) (D) \(60^{\circ}\)
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