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Chapter 20: Problem 72

The radius of curvature of the face of planoconvex lens is \(12 \mathrm{~cm}\) and its refractive index is \(1.5\). If the plane surface of the lens is now silvered, then the focal length of the lens is : (a) \(26 \mathrm{~cm}\) (b) \(22 \mathrm{~cm}\) (c) \(24 \mathrm{~cm}\) (d) \(20 \mathrm{~cm}\)

Short Answer

Expert verified
The focal length of the lens is 24 cm.

Step by step solution

01

Given data identification

We are given a planoconvex lens with a radius of curvature \( R = 12 \text{ cm} \) and a refractive index \( n = 1.5 \). The plane surface is silvered, thus converting it into a lens-mirror combination.
02

Effective refractive index

Since the lens is a planoconvex lens, its refractive index \( n = 1.5 \) will be used in the lens formula. Since the plane side is silvered, we'll consider using the formula for a mirror-lens combination effectively.
03

Using lens-maker's formula

We use the lens-maker's formula for a planoconvex lens:\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]For a planoconvex lens, \( R_1 = R = 12 \text{ cm} \) and \( R_2 = \infty \):\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{12} - \frac{1}{\infty} \right) = \frac{0.5}{12} = \frac{1}{24} \text{ cm}^{-1} \]Thus, \( f = 24 \text{ cm} \) for the lens alone.
04

Focal length of silvered lens

Since the plane side of the lens is silvered, the combination acts like a mirror. The effective focal length \( F \) of such a system can be found using:\[ \frac{1}{F} = \frac{1}{f_l} + \frac{2}{f_m} \]where \( f_m \) is the focal length of the mirror (plane mirror has infinite focal length).Thus, the focal length of the silvered side (as a mirror) will introduce no change:\[ \frac{1}{F} = \frac{1}{24} + \frac{2}{\infty} = \frac{1}{24} \text{ cm}^{-1} \]Therefore, the effective focal length is \( F = 24 \text{ cm} \).

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Key Concepts

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

Understanding the Radius of Curvature
The radius of curvature is a fundamental concept when dealing with lenses and mirrors. It is the distance between the center of the curvature of the lens surface and any given point on that surface. For a planoconvex lens, one side is convex with a tangible radius (in this case, 12 cm), while the other side is flat and has an infinite radius of curvature.
  • The radius of curvature helps to determine how the lens will bend light rays that pass through it.
  • It is an essential variable in calculating the focal length of a lens using the lens-maker's formula.
  • A smaller radius of curvature results in a more strongly curved lens, capable of bending light rays more sharply.

For a planoconvex lens like the one in the problem, you only need to consider the curvy side since the planar side has an infinite radius, simplifying the calculations.
The Role of Refractive Index
The refractive index is all about how much a material slows down and bends light. It's represented by the symbol \( n \), and in this context, it measures how light changes direction when entering or exiting the lens compared to passing through air. Here, the refractive index is 1.5 for the planoconvex lens.
  • The refractive index definition is simple: the speed of light in a vacuum divided by the speed of light in the material.
  • This concept is crucial for determining how lenses focus light easily, which impacts their focal length.
  • A higher refractive index means light refracts more and can result in a shorter focal length.
This makes the refractive index crucial in using the lens-maker's formula, which helps calculate the effective focal length of lenses efficiently.
Evaluating Focal Length
The focal length of a lens tells us where light rays converge after passing through it. It's a key factor in lens-making and is determined by both the radius of curvature and refractive index as seen in the lens-maker's formula. In this exercise, for a planoconvex lens with a refractive index of 1.5, the focal length is calculated to be 24 cm.
  • The formula used is \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), where \( R_2 \) is infinity.
  • This is simplified since \( \frac{1}{\infty} = 0 \), giving us the \( \frac{1}{f} = \frac{0.5}{12} \).
  • Thus, the focal length \( f = 24 \) cm for the lens itself.
For the silvered lens, it behaves as a mirror-lens combination, which does not change the effective overall focal length due to the infinite mirror effect, maintaining a value of 24 cm.

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Most popular questions from this chapter

A lens forms a sharp image of a real object on a screen. On inserting a parallel slide between the lens and the screen with its thickness along the principal axis of the lens, it is found necessary to shift the screen parallel to itself distance \(d\) away from the lens for getting image sharply focussed on it. If the refractive index of the glass relative to air is \(\mu\), the thickness of the slab is: (a) \(\frac{d}{\mu}\) (b) \(\mu d\) (c) \(\frac{r d}{\mu-1}\) (d) \((\mu-1) \frac{d}{\mu}\)

The electric permittivity and magnetic permeability of free space are \(\varepsilon_{0}\) and \(\mu_{0}\), respectively. The index of refraction of the medium, if \(\varepsilon\) and \(\mu\) are the electric permittivity and magnetic permeability in a medium is : (a) \(\frac{\varepsilon \mu}{\varepsilon_{0} \mu_{0}}\) (b) \(\left(\frac{\varepsilon \mu}{\varepsilon_{0} \mu_{0}}\right)^{1 / 2}\) (c) \(\frac{\varepsilon_{0} \mu_{0}}{\varepsilon \mu}\) (d) \(\left(\frac{\varepsilon_{0} \mu_{0}}{\varepsilon \mu}\right)^{1 / 2}\)

A thin prism of angle \(7^{\circ}\) made of glass of refractive index \(1.5\) is combined with another prism made of glass of \(\mu=1.75\) to produce dispersion without deviation. The angle of second prism is: (a) \(7^{\circ}\) (b) \(4.67^{\circ}\) (c) \(9^{\circ}\) (d) \(5^{\circ}\)

In the figure, a point source \(' P^{\prime}\) is placed at a height \(h\) above the plane mirror in a medium of refractive index \(\mu\). An observer \(O\), vertically above \(P\), outside the liquid, sees \(P\) and its image in the mirror. The apparent distance between these two is: (a) \(2 \mu h\) (b) \(\frac{2 h}{\mu}\) (c) \(\frac{2 h}{\mu-1}\) (d) \(h\left(1+\frac{1}{\mu}\right)\)

A compound microscope has an eye piece of focal length \(10 \mathrm{~cm}\) and an objective of focal length \(4 \mathrm{~cm}\). The magnification, if an object is kept at a distance of \(5 \mathrm{~cm}\) from the objective and final image is formed at the least distance of distinct vision \((20 \mathrm{~cm})\), is : (a) 10 (b) 11 (c) 12 (d) 13
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