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Chapter 23: Problem 75

An object is viewed through a compound microscope and appears in focus when it is \(5 \mathrm{~mm}\) away from the objective lens. When a sheet of transparent material \(3 \mathrm{~mm}\) thick is placed between the objective and the microscope, the objective lens has to be moved \(1 \mathrm{~mm}\) to bring to object back into the focus. The refraetive index of the transparent material is (a) \(1.5\) (b) \(1.6\) (c) \(1.8\) (d) \(2.0\)

Short Answer

Expert verified
The refractive index is 1.5, option (a).

Step by step solution

01

Understanding the Problem

We need to determine the refractive index of a transparent material inserted between an object and a microscope lens. Initially, the object is in focus at a distance of 5 mm. After inserting a 3 mm thick material, the lens must move 1 mm to refocus. The refractive index (n) can be found by analyzing these changes in focus.
02

Using Lens Equation

When adding the transparent material, the actual path of light is through the material, which affects the focus. The shift in image position due to refractive materials is given by the formula: \[d = t \left(1 - \frac{1}{n} \right)\]where \(d\) is the change in lens position (1 mm), \(t\) is the thickness of the material (3 mm), and \(n\) is the refractive index we need to find.
03

Substitute Known Values

Substitute the known values into the equation:\[1 = 3 \left(1 - \frac{1}{n}\right)\].This equation represents the lens needing to move 1 mm to refocus when a 3 mm transparent material is added.
04

Solve for Refractive Index

Solve the equation for \(n\):\[1 = 3 - \frac{3}{n}\]Rearrange to find \(n\):\[\frac{3}{n} = 3 - 1\]\[\frac{3}{n} = 2\]solve for \(n\):\[n = \frac{3}{2} = 1.5\].
05

Verify the Solution

Verify by considering if the calculated refractive index (1.5) makes physical sense with the placement and effect of the material. The extra 1 mm of movement accounting for a full 3 mm of material thickness is a typical refraction effect described by our final index.

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Key Concepts

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

Compound Microscope
A compound microscope is a powerful tool used to view tiny objects that can't be seen with the naked eye. It consists of two primary lenses: an objective lens and an eyepiece lens. Together, they magnify the specimen placed on the microscope stage.
  • The **objective lens** is closer to the object being observed and is responsible for the initial magnification of the image.
  • The **eyepiece lens** further magnifies the image for the viewer.
Using a compound microscope involves adjusting the distance between these lenses and the specimen to bring it into sharp focus. This focus adjustment becomes particularly interesting when a transparent material is introduced, as it influences how light reaches and magnifies the sample.
Optical Path Length
Optical path length is a term used to describe how light travels through different media before reaching the observer's eye. It is a product of the actual physical path the light travels and the refractive index of the material.
When you introduce a transparent material between the objective lens and the specimen, the path of light bends due to refraction. This bending is dependent on the refractive index of the material, altering the optical path length.
For a microscope, the correct optical path length must be maintained to ensure the object remains in focus. If the path is altered, adjustments must be made, such as moving the objective lens. In the given exercise, the lens needed to be moved by 1 mm to regain focus after a 3 mm thick transparent material was added. Such calculations are based on the relationship between the optical path length and the refractive index.
Transparent Material
A transparent material allows light to pass through with minimal scattering. Examples include glass and clear plastics. The exercise introduces a transparent material of 3 mm thickness.
The key property of transparent materials in this context is their refractive index, which affects how much light bends as it travels through the material.
  • This bending, or refraction, happens because light travels at different speeds in different materials.
  • In our exercise, inserting the transparent material required focus adjustments, since the refractive index determined how significantly the light's path was altered.
By understanding the refractive index, you can predict how much the light path will change and how to adjust the microscope accordingly.
Focus Adjustment
Focus adjustment in a microscope is crucial for bringing details into clear view. When objects are properly focused, they appear sharp and detailed through the lenses.
Introducing a transparent material changes the travel path of the light, which means that the previously set focus might no longer be correct. Therefore, adjustments are necessary to bring the image back into clarity.
  • In our exercise, this was achieved by moving the objective lens by 1 mm.
  • This adjustment is directly related to the refractive index of the material added, which in this case was calculated to be 1.5.
Mastering focus adjustment ensures that changes in the setup, such as inserting different materials, do not lead to blurred or unclear images.

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Most popular questions from this chapter

A plane mirror is approaching you at \(10 \mathrm{cms}^{-1}\). Your image shall approach you with a speed of (a) \(+10 \mathrm{cms}^{-1}\) (b) \(-10 \mathrm{cms}^{-1}\) (c) \(+20 \mathrm{cms}^{-1}\) (d) \(-20 \mathrm{cms}^{-1}\)

A small bulb is placed at the bottom of a tank containing water to a depth of \(80 \mathrm{~cm}\). What is the area of the surface of water through which light from the bulb ean emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) [NCERT] (a) \(4.6 \mathrm{~m}^{2}\) (b) \(3.2 \mathrm{~m}\) (c) \(5.6 \mathrm{~m}^{2}\) (d) \(2.6 \mathrm{~m}^{2}\)

A convex mirror forms an image one-fourth the size of the object. If object is at a distance of \(0.5 \mathrm{~m}\) from the mirror, the focal length of mirror is (a) \(0.17 \mathrm{~m}\) (b) \(-1.5 \mathrm{~m}\) (c) \(0.4 \mathrm{~m}\) (d) \(-0.4 \mathrm{~m}\)

A car is moving with at a constant speed of \(60 \mathrm{kmh}^{-1}\) on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at distance of \(100 \mathrm{~m}\) and is approaching with a speed of \(5 \mathrm{~km} \mathrm{~h}^{-1}\). In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every \(2 \mathrm{~s}\) till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct? (a) The speed of the car in the rear is \(65 \mathrm{~km} \mathrm{~h}^{-1}\) (b) In the side mirror the car in the rear would appear to approach with a speed of \(5 \mathrm{~km} \mathrm{~h}^{-1}\) to the driver of the leading car (c) In the rear view minor, the speed of the approaching car would appeat to decrease as the distance between the. cars decreases (d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases

A ray of light falls on a transparent glass slab with refractive index (relative to air) of \(1.62 .\) The angle of incidence for which the reflected and refracted rays are mutually perpendicular is (a) \(\tan ^{-1}(162)\) (b) \(\sin ^{-1}(162)\) (c) \(\cos ^{-1}(162)\) (d) None of these
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