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Chapter 24: Problem 42

A lens having focal length \(f\) and aperture of diameter \(\bar{d}\) forms an image of intensity \(I\). Aperture of diameter \(\frac{d}{2}\) in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively (a) \(f\) and \(\frac{I}{4}\) (b) \(\frac{3 f}{4}\) and \(\frac{I}{2}\) (c) \(f\) and \(\frac{3 I}{4}\) (d) \(\frac{f}{2}\) and \(\frac{I}{2}\)

Short Answer

Expert verified
(c) \( f \) and \( \frac{3I}{4} \).

Step by step solution

01

Understanding the Setup

You have a lens with a focal length \( f \) and an aperture of diameter \( \bar{d} \). Initially, the full aperture is used to form an image of intensity \( I \).
02

Effect of Covering Part of the Aperture

Part of the lens's aperture, up to a diameter of \( \frac{\bar{d}}{2} \), is covered. This corresponds to covering \( \frac{1}{4} \) of the full area of the aperture, reducing the area to \( \frac{3}{4} \) of its original size.
03

Impact on Focal Length

The focal length of the lens, \( f \), is a property of the lens and does not change due to partial covering of the aperture. Therefore, the focal length remains \( f \).
04

Calculating New Intensity

The intensity of the image relates to the area of the lens being used. Since only \( \frac{3}{4} \) of the aperture is now used, the intensity reduces to \( \frac{3}{4} \times I \).
05

Solution Conclusion

The focal length stays the same at \( f \), and the new intensity is \( \frac{3}{4} \) of the original intensity \( I \). Therefore, the lens focal length and intensity of the image now are \( f \) and \( \frac{3I}{4} \) respectively, which corresponds to option (c).

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Key Concepts

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

Understanding Focal Length
The focal length of a lens is the distance from the lens to the point where it brings light rays into focus. It determines how strongly a lens converges or diverges light. For converging lenses, the focal length is positive, while for diverging lenses, it鈥檚 negative. This concept is crucial in optics because:
  • It affects the magnification of the image.
  • Shorter focal lengths yield wider fields of view, whereas longer focal lengths offer greater magnification.
In the presented problem, the focal length of the lens remains unchanged even when part of its aperture is covered. This is because the focal length is an inherent property of the lens, dependent on its shape and the material from which it is made.
Exploring Lens Aperture
The lens aperture pertains to the opening through which light enters a lens. It is typically circular, and its size plays a crucial role in the brightness and depth of field of the resulting image. Larger apertures let in more light, producing brighter images and reduced depth of field, which is effective for isolating subjects. In the given physics problem, initially, you have a full aperture, yet when a central region is blocked:
  • The effective area of the aperture is reduced.
  • Less light is permitted through, impacting image intensity.
Here, the aperture reduced from \(ar{d}\) to \(\frac{\bar{d}}{2}\), effectively decreasing the useable aperture area to \(\frac{3}{4}\) of its original size, which directly affects the image's intensity.
Image Intensity Insights
Image intensity generally describes the brightness or strength of the image formed by a lens. It is directly related to the amount of light passing through the lens's aperture. More open apertures allow more light, leading to higher intensity. When portions of the aperture are obstructed, such as in the exercise given,:
  • The intensity reduces proportional to the area exposed.
  • Image appears less bright because less light is concentrated in the image location.
In this scenario, covering the \(\frac{\bar{d}}{2}\) diameter lens aperture leads to only \(\frac{3}{4}\) of the light being used, reducing the intensity to \(\frac{3I}{4}\).
Explaining Light Intensity
Light intensity refers to the energy of light that is emitted or absorbed by a surface area. It's measured in units like lux or foot-candles and is a vital concept in optics. Factors influencing light intensity include:
  • The source's power and distance from an object.
  • The aperture size allowing light to enter and pass through a lens.
In this optical problem, the aperture size was reduced, effectively lowering the light intensity incident on the image plane. Since only \(\frac{3}{4}\) of the original aperture is utilized, the corresponding light intensity reaching the image is also reduced accordingly.
Solving Physics Problems in Optics
Physics problems involving lenses require understanding key optical principles such as focal length, aperture size, and how they influence both image quality and intensity. When solving these problems:
  • Start by identifying known quantities like lens properties and initial conditions.
  • Understand how light interacts with the lens and how changes to its structure, like aperture reduction, affect the outcome.
  • Apply relevant equations and physics laws to analyze the given situation.
Analyzing and manipulating such parameters reveals the intricate nature of optics, where small adjustments can lead to significant changes in an image's characteristics.

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Most popular questions from this chapter

A convex spherical refracting surface with radius \(R\) separates a medium having refractive index \(5 / 2\) from air. As an object is moved towards the surface from far away from the surface along the central axis, its image (a) changes from real to virtual when it is at a distance \(R\) from the surface (b) changes from virtual to real when it is at a distance \(R\) from the surface (c) changes from real to virtual when it is at a distance ( \(2 R / 3\) ) from the surface (d) changes from virtual to real when it is at a distance ( \(2 R / 3\) ) from the surface

A telescope of diameter \(2 \mathrm{~m}\) uses light of wavelength \(5000 \AA\) for viewing stars. The minimum angular separation between two stars whose image is just resolved by this telescope, is (a) \(4 \times 10^{-4} \mathrm{rad}\) (b) \(0.25 \times 10^{-6} \mathrm{rad}\) (c) \(0.31 \times 10^{-6} \mathrm{rad}\) (d) \(5.0 \times 10^{-3} \mathrm{rad}\)

The least distance of distinct vision for a long-sighted person is \(x \mathrm{~m}\). He wants to read a newspaper placed at a distance \((x / 2) \mathrm{m} .\) The power of the spectacle lens used by him is \(x D\). What is the value of \(x\) ? (a) \(0.5\) (b) \(1.0\) (c) \(1.5\) (d) \(2.0\)

Rainbow is formed due to a combination of (a) dispersion and total internal reflection (b) refraction and absorption (c) dispersion and focussing (d) refraction and scattering

A point object is placed at a distance of \(20 \mathrm{~cm}\) from a thin plano- convex lens of focal length \(15 \mathrm{~cm}\). If the plane surface is silvered, the image will form at (a) \(60 \mathrm{~cm}\) from left of \(A B\) (b) \(30 \mathrm{~cm}\) from left of \(A B\) (c) \(12 \mathrm{~cm}\) from left of \(A B\) (d) \(60 \mathrm{~cm}\) from right of \(A B\)
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