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Two mirrors each \(1.6 \mathrm{~m}\) long are facing each other. The distance between the mirrors is \(20 \mathrm{~cm}\). A light ray is incident on one end of the mirror at an angle of incidence of \(30^{\circ} .\) How many times is the ray reflected before it reaches the other end?

Short Answer

Expert verified
The ray is reflected 4 times before reaching the other end.

Step by step solution

01

Convert Units

First, convert all the lengths to the same unit. The distance between the mirrors is given in centimeters, so let's convert 1.6 meters to centimeters. We have:\[ 1.6 ext{ m} = 160 ext{ cm} \]
02

Calculate the Path

The light ray travels back and forth between the mirrors at a 30-degree angle. The path forms a triangle with the base inside the mirrors and the hypotenuse being the path of the light. Given the symmetry and angle, each complete reflection in this zig-zag path results in a vertical shift equal to:\[ x = 2 imes 20 imes an(30^ ext{o}) \]where the factor of 2 accounts for the round trip between the mirrors. This calculation finds the distance covered per round trip between mirrors.
03

Compute Vertical Shift

Find the total vertical distance after each pair of reflections, using the formula for the vertical shift:\[ x = 2 imes 20 ext{ cm} imes rac{1}{ an(30^ ext{o})} \]\[ an(30^ ext{o}) ext{ is } rac{1}{ an(30^ ext{o})} = rac{1}{ rac{ ext{Opposite}}{ ext{Adjacent}}} = rac{1}{ rac{1}{ an(30^ ext{o})}} ext{ which simplifies to } \sqrt{3} \] Hence, \[ x = 2 \times 20 \times \sqrt{3} = 40 \sqrt{3} \] cm.
04

Count Reflections

Count the number of vertical shifts required to reach the end of the mirror. If each move results in a vertical shift of 40 cm (height of the mirror), then the light must move up the entire length of the mirror, which is 160 cm:\[ n = \frac{160}{40} = 4 \]Thus, the ray reflects 4 times before reaching the end of the mirror.

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Key Concepts

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

Angle of Incidence
The angle of incidence is the angle between the incoming light ray and the normal (an imaginary line perpendicular to the surface) at the point of contact on the mirror. In this exercise, the light ray hits the first mirror at an angle of incidence of \(30^\circ\). This angle is crucial because it determines how the light will behave as it meets the mirror surface. According to the law of reflection, the angle of incidence is equal to the angle of reflection. Therefore, after hitting the surface, the light ray will reflect off at the same \(30^\circ\) angle as it enters. This predictable behavior helps us calculate the path and the number of reflections the light ray will undergo.
Path of Light
The path of a light ray is the trajectory it follows after striking a reflective surface. In this scenario, the light ray bounces back and forth between two mirrors, creating a zig-zagging pattern. This pattern occurs because each interaction with the mirror surfaces follows the law of reflection. Imagining a series of tiny, identical triangles formed by each bounce helps us understand this path more clearly.
  • The base of the triangle is equal to the distance between the mirrors (20 cm).
  • The hypotenuse represents the distance the light ray actually travels as it moves from one mirror to the other and back again.
  • The vertical component or shift is determined by the angle of incidence, affecting how far the ray moves up or down the mirror with each reflection.
Understanding the path of light is crucial for calculating how reflections bring the ray to the opposite end of the mirror.
Tangent Function
The tangent function is a trigonometric function that relates the angles of a triangle to the ratios of its sides. In our example, the tangent of \(30^\circ\) plays a critical role in determining how far the light travels vertically with each reflection. The tangent of an angle is defined as the ratio of the opposite side to the adjacent side in a right triangle:\[\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}}\]Since we know the distance between the mirrors (the base of each triangle) and the angle, we can use the tangent function to calculate the height shift (vertical movement) each time the light reflects. Substituting values:\[x = 2 \times 20 \times \tan(30^\circ)\]This equation gives us the key value of the vertical movement by relying on tangent, showcasing why this function is useful for problems involving angles and distances.
Unit Conversion
Unit conversion is a method used to ensure all measurements are in the same units, which is essential for accurate calculations. In this exercise, the length of the mirrors is given in meters (1.6 m), while the distance between them is in centimeters (20 cm). To simplify calculations, we convert the length of the mirrors from meters to centimeters. This way, all calculations can be performed more straightforwardly and consistently. Here's how it is done:
  • Length in meters: 1.6 m
  • Conversion to centimeters: \(1.6 \text{ m} = 160 \text{ cm}\)
Converting these measures allows us to directly compare and compute values, ensuring nothing confuses or mismatches during the calculation process. This step is often one of the first done in a mathematics problem to set up a coherent and unified problem-solving environment.

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Most popular questions from this chapter

A bullet of mass \(m_{2}\) is fired from a gun of mass \(m_{1}\) with horizontal velocity \(v .\) A plane mirror is fixed at gun facing towards bullet. The velocity of the image of bullet formed by the plane mirror with respect to bullet is: (a) \(\left(1+\frac{m_{2}}{m_{1}}\right)\) (b) \(\left(\frac{m_{1}+m_{2}}{m_{1}}\right) v\) (c) \(\frac{2\left(m_{1}+m_{2}\right)}{m_{1}} v\) (d) none of these

A point object \(P\) is placed at centre of curvature of a concave mirror of focal length \(25 \mathrm{~cm}\). The mirror is cut into two halves and shifted symmetrically \(1 \mathrm{~cm}\) apart in perpendicular to the optical axis. The distance between images formed by both parts is: (a) \(2 \mathrm{~cm}\) (b) \(1 \mathrm{~cm}\) (c) \(3 \mathrm{~cm}\) (d) \(4 \mathrm{~cm}\)

An object of height \(5 \mathrm{~cm}\) is placed in midway between a concave mirror of radius of curvature \(30 \mathrm{~cm}\) and a convex mirror of radius of curvature \(30 \mathrm{~cm}\). The mirrors are placed opposite to each other and are \(60 \mathrm{~cm}\) apart. The position of the image formed by reflection at convex mirror is : (a) \(10 \mathrm{~cm}\) (b) \(20 \mathrm{~cm}\) (c) \(15 \mathrm{~cm}\) (d) \(13 \mathrm{~cm}\)

A ray is incident on a plane surface. If \(\hat{i}+\hat{j}-\hat{k}\) represents a vector along the direction of incident ray. \(\hat{i}+\hat{j}\) is a vector along normal on incident point in the plane of incident and reflected ray. Then vector along the direction of reflected ray is: (a) \(-\frac{1}{\sqrt{19}}(-3 \hat{\mathbf{i}}+3 \hat{j}+\mathbf{k})\) (b) \(\frac{1}{\sqrt{19}}(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\cdot \hat{\mathrm{k}})\) (c) \(-\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\) (d) \(\hat{\mathbf{k}}\)

When an object is placed at a distance of \(25 \mathrm{~cm}\) from a mirror, the magnification is \(m_{1}\). But the magnification becomes \(m_{2}\), when the object is moved \(15 \mathrm{~cm}\) farther away with respect to the earlier position. If \(\frac{m_{1}}{m_{2}}=4\), then find the focal length of the mirror and what type of mirror it is? (a) \(20 \mathrm{~cm}\), convex (b) \(20 \mathrm{~cm}\), concave (c) \(10 \mathrm{~cm}\), convex (d) \(10 \mathrm{~cm}\), concave

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