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Chapter 23: Problem 59

The radius of the convex surface of plane-convex lens is \(20 \mathrm{~cm}\) and the refractive index of the material of the lens is 1.5. The focal length of the lens is (a) \(30 \mathrm{~cm}\) (b) \(50 \mathrm{~cm}\) (c) \(20 \mathrm{~cm}\) (d) \(40 \mathrm{~cm}\)

Short Answer

Expert verified
The correct answer is (d) 40 cm.

Step by step solution

01

Understand Lens Formula

For a lens, the formula relating the focal length \( f \), the refractive index \( n \), and the radius of curvature \( R \) of the lens is given by the lens maker's equation: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]. For a plane-convex lens, one side is flat, so \( R_2 = \infty \), simplifying to \[ \frac{1}{f} = (n - 1) \frac{1}{R} \]. We need to calculate the focal length \( f \).
02

Plug in Known Values

In the simplified lens maker's formula \( \frac{1}{f} = (n-1) \frac{1}{R} \), plug in \( n = 1.5 \) and \( R = 20 \text{ cm} \). This gives us: \( \frac{1}{f} = (1.5 - 1) \frac{1}{20} \).
03

Solve for Focal Length

Calculate the change in refractive index by subtracting 1 from \( n \): \( 1.5 - 1 = 0.5 \). Therefore, the equation becomes \( \frac{1}{f} = 0.5 \times \frac{1}{20} \). Simplify this to get \( \frac{1}{f} = \frac{0.5}{20} = \frac{1}{40} \).
04

Calculate Final Answer

Now solve for \( f \) by taking the reciprocal of both sides: \( f = 40 \text{ cm} \).

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Key Concepts

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

Lens Maker's Equation
The lens maker's equation is fundamental in optics. It establishes a relationship between the focal length of a lens, its refractive index, and the curvature of its surfaces. For any lens, this equation is expressed as: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Where:
  • \( f \) is the focal length, the distance where parallel rays converge after passing through the lens.
  • \( n \) stands for the refractive index, a measure of how much light is bent or refracted when entering the lens.
  • \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens.
For a plane-convex lens, like the one in our problem, one side is flat, making its radius of curvature \( R_2 \) infinite. Consequently, the equation simplifies to \[ \frac{1}{f} = (n - 1) \frac{1}{R} \]. This simplification makes it easier to calculate lenses' focal lengths where one surface is planar.
Focal Length
The focal length of a lens is a critical concept in optics. It is defined as the distance from the lens at which parallel rays of light converge. This fundamental distance determines how a lens focuses light and, subsequently, images. The focal length is influenced by:
  • Lens curvature: Highly curved lenses have shorter focal lengths and converge light more rapidly.
  • Refractive index: Lenses made from materials with a higher refractive index can also shorten the focal length.
  • Lens maker's equation: As seen, the equation directly links these variables to determine \( f \).
In practical usage, a shorter focal length means a wider field of view, while longer focal lengths narrow the field, focusing more on distant objects.
Refractive Index
The refractive index is a number that describes how fast light travels through a material. It is an essential parameter in optical systems, greatly influencing lens behavior. Formally defined, it is the ratio of the speed of light in a vacuum to its speed in the material. This is given by: \[ n = \frac{c}{v} \] where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light within the material.Several factors influence the refractive index:
  • Material composition: Denser materials generally have higher refractive indices.
  • Wavelength of light: The refractive index can vary with the wavelength due to dispersion, meaning it is different for various colors.
The refractive index is vital for determining how light is bent and focused by lenses, affecting their focal length based on the lens maker鈥檚 equation.

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Most popular questions from this chapter

At what angle should a ray of light be incident on the face of a prism of refracting angle \(60^{\circ}\) so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is \(1.524\). [NCERT] (a) \(16^{\circ}\) (b) \(29^{\circ}\) (c) \(45^{\circ}\) (d) \(58^{\circ}\)

The magnifying power of an astronomical telescope is 10 and the focal length of its eye-piece is \(20 \mathrm{~cm}\). The focal length of its objective will be (a) \(200 \mathrm{~cm}\) (b) \(2 \mathrm{~cm}\) (c) \(0.5 \mathrm{~cm}\) (d) \(0.5 \times 10^{-2} \mathrm{~cm}\)

To focal length of a concave mirror is \(12 \mathrm{~cm}\). Where should an object of length \(4 \mathrm{~cm}\) be placed so that an image \(1 \mathrm{~cm}\) long is formed? (a) \(48 \mathrm{~cm}\) (b) \(3 \mathrm{~cm}\) (c) \(-60 \mathrm{~cm}\) (d) \(15 \mathrm{~cm}\)

A concave lens with unequal radii of curvature made of glass \(\left(\mu_{B}=15\right)\) has focal length of \(40 \mathrm{~cm}\). If it is immersed in a liquid of refractive index \(\mu=2\), then (a) it behaves like a convex lens of \(80 \mathrm{~cm}\) focal length (b) it behaves like a concave lens of \(20 \mathrm{~cm}\) focal length (c) its focal length becomes \(60 \mathrm{~cm}\) (d) nothing can be said

The power of a thin convex lens \(\left(_{a} n_{g}=1.5\right)\) is \(+5.0 \mathrm{D}\). When it is placed in a liquid of refractive index \(_{a} n_{e}\), then it behaves as a concave lens of focal length \(100 \mathrm{~cm}\). The refractive index of the liquid \(_{a} n_{l}\) will be (a) \(5 / 3\) (b) \(4 / 3\) (c) \(\sqrt{3}\) (d) \(5 / 4\)
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