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\(P\) is a point on the axis of a concave mirror. The image of \(P\) formed by the mirror, coincides with \(P . \mathrm{A}\) rectangular glass slab of thiekness \(t\) and refractive index \(\mu\) is now introduced between \(P\) and the mirror. For image of \(P\) to coincide with \(P\) again, the mirror must be moved (a) towards \(P\) by \((\bar{\mu}-\mathbf{I}) t\) (b) away from \(P\) by \((\mu-1) t\) (c) towards \(P\) by \(t\left(1-\frac{1}{\mu}\right)\) (d) away from \(P t\left(1-\frac{1}{\mu}\right)\)

Short Answer

Expert verified
The mirror must be moved away from \( P \) by \((\mu-1)t\). (Option b)

Step by step solution

01

Understand the Problem

We have a point \( P \) on the axis of a concave mirror whose image coincides with the point \( P \). When a glass slab with thickness \( t \) and refractive index \( \mu \) is placed between \( P \) and the mirror, we need to find out how the mirror should be moved for the image to still coincide with \( P \).
02

Effect of Glass Slab

The introduction of a glass slab changes the optical path of light. The optical path inside the glass slab of thickness \( t \) is effectively altered by a factor of \( \mu \), making the light path effectively longer than it is in the air. This altered path needs to be compensated by moving the mirror.
03

Calculate the Change in Optical Path

The optical path difference caused by the slab is given by \((\mu - 1) t\), where \( t \) is the thickness of the slab and \( \mu \) is its refractive index. This is because light travels the same number of waves in a shorter actual distance inside the slab.
04

Determine Mirror Movement Direction

Since the optical path is effectively increased by \((\mu - 1) t\), to make the image coincide with \( P \) again, the real path must be increased. Therefore, the mirror needs to be moved away from \( P \) by \((\mu - 1) t\).
05

Determine the Correct Option

Given the movements described, the correct choice that matches the necessary movement of the mirror is option (b): "away from \( P \) by \((\mu-1) t\)."

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Key Concepts

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

Optical Path Difference
The optical path difference is an essential concept in optics when dealing with mediums other than air. It describes how the path of light changes when it encounters materials with different refractive indices. When light travels through a medium, such as a glass slab, it slows down compared to its speed in a vacuum or air. This slowing effect causes the light to take longer to pass through the medium. Although the physical distance may remain constant, the optical path length—how the light "senses" its journey—becomes longer. In simpler terms, if you think of light as a wave, the number of wave crests that fit into a distance changes when the medium changes. This difference is captured as the optical path difference. It plays a significant role when aligning optical systems since it determines whether light waves will interfere constructively or destructively upon exiting the medium or interacting with surfaces like mirrors.
Refractive Index
The refractive index, often represented by the symbol \(\mu\), is a measure of how much the speed of light is reduced inside a material compared to its speed in a vacuum. It's a crucial factor in determining how light behaves when entering different substances. The refractive index is a dimensionless number, and for most transparent materials, it's greater than 1. When light moves from a medium of lower refractive index (like air) into a medium of higher refractive index (such as glass), it slows down and bends toward the normal line at the boundary, an effect described by Snell's Law:\[ n_1 \sin \theta_1 = n_2 \sin \theta_2\] Where \( n_1\) and \(n_2\) are the refractive indices of the initial and final mediums, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively. This bending of light is what gives lenses their magic, enabling us to focus and manipulate light in various ways.
Glass Slab Refraction
When light travels through a glass slab, it experiences refraction at each boundary it encounters. First, as the light enters the glass from the air, it bends towards the normal due to the higher refractive index of glass compared to air. This bending continues when light exits the glass slab into the air, where it bends away from the normal.Despite these changes, if the slab is uniform, the light eventually emerges parallel to its original path. However, due to the optical path difference caused by the slab's refractive index, the effective path becomes longer. The effective optical path length is represented by the equation: \[ ( ext{optical path length}) = \mu \times t\]Where \(\mu\) is the refractive index and \(t\) is the thickness of the slab. This longer path needs consideration, especially when the slab is placed in systems relying on precise optical alignment, such as imaging with mirrors or lenses. Understanding how to compensate for this extended path ensures systems maintain their accuracy and functionality.

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Most popular questions from this chapter

One of the refracting surfaces of a prism of angle \(30^{\circ}\) is silvered. A ray of light incident at an angle of \(60^{\circ}\) retraces its path. The refractive index of the material of prism is (a) \(\sqrt{3}\) (b) \(3 / 2\) (c) 2 (d) \(\sqrt{2}\)

For a normal eye, the far point is at infinity and the near point of distinct vision is about \(25 \mathrm{~cm}\) in front of the eye. The cornea of the eye provides a converging power of about \(40 \mathrm{D}\), and the least converging power of the eye lens behind the cornea is about \(20 \mathrm{D}\). From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye lens) of a normal eye. [al 10 tol4 D (b) 20 to \(24 \mathrm{D}\) (c) 28 to \(32 \mathrm{D}\) (d) 14 to \(18 \mathrm{D}\)

A plano-convex lens has a thickness of \(4 \mathrm{~cm}\). When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be \(3 \mathrm{~cm}\). If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be \(25 / 8 \mathrm{~cm}\). Find the foeal length of the lens. Assume thickness to be negligible (a) \(85 \mathrm{~cm}\) (b) \(59 \mathrm{~cm}\) (c) \(75 \mathrm{~cm}\) (d) \(7.5 \mathrm{~cm}\)

A plane mirror is reflecting a ray of incident light is rotated through an angle of about an axis through the point of incidence in the plane of the mirror perpendicular to the plane of incident, then (a) the reflected ray rotates through an angle \(2 \theta\) (b) the reflected ray rotates through an angle of \(\theta\) (c) the reflected ray does not rotate (d) None of the above

Two convex lenses placed in contaet form the image of a distance object at \(P .\) If the lens \(B\) is moved to the right, the image will (a) move to the left (b) move to the right (c) remain at \(P\) (d) move either to the left or right, depending upon focal Icngth of the lenses

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