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Chapter 19: Problem 28

A narrow beam of light after reflection by a plane mirror falls on a scale \(100 \mathrm{~cm}\) from the mirror. When the mirror is rotated a little the spot moves through \(2 \mathrm{~cm}\), the angle, through which the mirror is rotated is : (a) \(0.02 \mathrm{rad}\) (b) \(0.01 \mathrm{rad}\) (c) \(200 \mathrm{rad}\) (d) \(\frac{0.01}{180} \pi \mathrm{rad}\)

Short Answer

Expert verified
(b) \(0.01 \mathrm{rad}\)

Step by step solution

01

Understand the Problem

The problem involves a mirror reflecting light to a spot 100 cm away. When the mirror rotates, the reflected spot moves by 2 cm. We need to find the angle by which the mirror is rotated.
02

Visualize the Reflection

When a mirror rotates by an angle \( \theta \), the reflected ray rotates by \( 2\theta \). This is due to the law of reflection: the angle of incidence equals the angle of reflection. Therefore, any small change in the angle of the mirror results in a double change in the reflection angle.
03

Set Up the Trigonometric Equation

Consider a small angle \( \theta \) by which the mirror is rotated. The arc length by which the spot moves on the scale is 2 cm, and this corresponds to the angle \( 2\theta \) at a radius of 100 cm (the distance from the mirror).
04

Use Small Angle Approximation

For small angles, the arc length \( s \) is approximately equal to the radius \( r \) times the angle in radians \( \theta \): \[ s = r \cdot \theta \]Substitute the known values: \[ 2 = 100 \cdot 2\theta \]
05

Solve for \( \theta \)

From the equation \( 2 = 100 \cdot 2\theta \), solve for \( \theta \):\[ 2 = 200\theta \] \[ \theta = \frac{2}{200} \]\[ \theta = 0.01 \]
06

Conclusion

The angle through which the mirror is rotated is \( 0.01 \) radians.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angle of Incidence
In the context of a plane mirror, the angle of incidence is a fundamental concept. It refers to the angle formed between the incoming beam of light and the normal to the surface of the mirror. The normal is an imaginary line perpendicular to the surface of the mirror at the point where the light beam strikes it. Understanding this angle is crucial because it directly influences the behavior of the reflected light.

There are several important points to note about the angle of incidence:
  • It is always measured with respect to the normal, not the surface of the mirror.
  • The angle of incidence is equal to the angle of reflection. This is the core of the law of reflection, which means if a light beam strikes the mirror at a certain angle, it will be reflected off at the same angle.
  • This equality is key in predicting where a reflected light beam will end up, as demonstrated in the solved problem where the light spot moves based on the mirror's movement.
By understanding the angle of incidence, we can predict how light interacts with surfaces like mirrors, which is essential in many optical applications.
Small Angle Approximation
The small angle approximation is a useful mathematical concept often applied in optics and physics. It simplifies the calculations involving trigonometric functions by approximating angles measured in radians.

Here鈥檚 how and why it works in our example:
  • When an angle is small, its sine, tangent, and the angle itself in radians are almost equal. This makes calculations involving these angles much simpler.
  • In our mirror problem, the small angle approximation allows us to assume that the arc length (movement of the light spot) is simply the product of the radius (distance from mirror to point) and the angle (in radians) of displacement.
  • Therefore, the equation becomes straightforward: the arc length equals the radius times the angle, i.e., \( s = r \cdot \theta \), which was used to find the angle \( \theta \) when given the movement of the spot, demonstrating its practical utility.
This approximation is particularly handy in optics, where precise predictions of very small angular changes are needed, like in our mirror rotation scenario.
Law of Reflection
The law of reflection is a central tenet of optics, dictating the behavior of light when it encounters reflective surfaces like mirrors. It states that the angle of incidence is equal to the angle of reflection.

Let's break down this principle:
  • The law of reflection applies to all types of mirrors, whether flat (like plane mirrors) or curved.
  • In a plane mirror, as in the problem, this law ensures that any angle change of the mirror results in a predictable double angle change in the reflection, as seen when the mirror's slight movement caused a 2 cm shift in the light spot.
  • By applying this law, we were able to determine the relationship between the mirror's rotation and the reflected light path, crucial for solving the exercise problem accurately.
Understanding and applying the law of reflection is vital for designing optical devices, such as periscopes and telescopes, and everyday applications like using mirrors effectively.

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Most popular questions from this chapter

A car is moving towards a plane mirror at a speed of \(30 \mathrm{~m} / \mathrm{s}\). Then the relative speed of its image with respect to the car will be : (a) \(30 \mathrm{~m} / \mathrm{s}\) (b) \(60 \mathrm{~m} / \mathrm{s}\) (c) \(15 \mathrm{~m} / \mathrm{s}\) (d) \(45 \mathrm{~m} / \mathrm{s}\)

A clock fixed on a wall shows time \(04: 25: 37 .\) What time will its image in a plane mirror show? (a) \(07: 43: 32\) (b) \(07: 43: 32\) (c) \(07: 35: 23\) (d) \(43: 27: 36\)

A fluorescent lamp of length \(1 \mathrm{~m}\) is placed horizontal at a depth of \(1.2 \mathrm{~m}\) below a ceiling. A plane mirror of length \(0.6 \mathrm{~m}\) is placed below the lamp parallel to and symmetric to the lamp at a distance \(2.4 \mathrm{~m}\) from it. The length of the reflected patch of light on the ceiling is: (a) \(3 \mathrm{~m}\) (b) \(4 \mathrm{~m}\) (c) \(7 \mathrm{~m}\) (d) none of these

A plane mirror is placed in \(y-z\) plane facing towards negative \(x\) -axis. The mirror is moving parallel to \(y\) -axis with a speed of \(5 \mathrm{~cm} / \mathrm{s}\). A point object \(P\) is moving infront of the mirror with a velocity \(3 \hat{i}+4 \hat{j}\). The velocity of image is: (a) \(-3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\) (b) \(3 \hat{i}-4 \hat{j}\) (c) \(-3 \hat{i}\) (d) \(3 \hat{i}+4 \hat{j}\)

A pole \(5 \mathrm{~m}\) high is situated on a horizontal surface. Sun rays are incident at an angle \(30^{\circ}\) with the vertical. The size of shadow on horizontal surface is : (a) \(5 \mathrm{~m}\) (b) \(\frac{5}{\sqrt{3}}=\mathrm{m}\) (c) \(\frac{10}{\sqrt{3}} \mathrm{~m}\) (d) none of these
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