NEET Exam  >  NEET Tests  >  Topic-wise MCQ Tests for NEET  >  Case Based Questions Test: Ray Optics & Optical Instruments - 2 - NEET MCQ

Case Based Questions Test: Ray Optics & Optical Instruments - 2 - NEET MCQ


Test Description

15 Questions MCQ Test Topic-wise MCQ Tests for NEET - Case Based Questions Test: Ray Optics & Optical Instruments - 2

Case Based Questions Test: Ray Optics & Optical Instruments - 2 for NEET 2024 is part of Topic-wise MCQ Tests for NEET preparation. The Case Based Questions Test: Ray Optics & Optical Instruments - 2 questions and answers have been prepared according to the NEET exam syllabus.The Case Based Questions Test: Ray Optics & Optical Instruments - 2 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Case Based Questions Test: Ray Optics & Optical Instruments - 2 below.
Solutions of Case Based Questions Test: Ray Optics & Optical Instruments - 2 questions in English are available as part of our Topic-wise MCQ Tests for NEET for NEET & Case Based Questions Test: Ray Optics & Optical Instruments - 2 solutions in Hindi for Topic-wise MCQ Tests for NEET course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Case Based Questions Test: Ray Optics & Optical Instruments - 2 | 15 questions in 30 minutes | Mock test for NEET preparation | Free important questions MCQ to study Topic-wise MCQ Tests for NEET for NEET Exam | Download free PDF with solutions
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 1

Read the following text and answer the following questions on the basis of the same:

Negative Refractive Index: One of the most fundamental phenomena in optics is refraction. When a beam of light crosses the interface between two different materials, its path is altered depending on the difference in the refractive indices of the materials. The greater the difference, the greater the refraction of the beam. For all known naturally occurring materials the refractive index assumes only positive values. But does this have to be the case?

In 1967, Soviet physicist Victor Veselago hypothesized that a material with a negative refractive index could exist without violating any of the laws of physics.

Veselago predicted that this remarkable material would exhibit a wide variety of new optical phenomena. However, until recently no one had found such a material and Veselago’s ideas had remained untested. Recently, meta-material samples are being tested for negative refractive index. But the experiments show significant losses and this could be an intrinsic property of negativeindex materials.

Snell’s law is satisfied for the materials having a negative refractive index, but the direction of the refracted light ray is ‘mirror-imaged’ about the normal to the surface.

There will be an interesting difference in image formation if a vessel is filled with “negative water” having refractive index – 1.33 instead of regular water having refractive index 1.33.

Say, there is a fish in a vessel filled with negative water. The position of the fish is such that the observer cannot see it due to normal refraction since the refracted ray does not reach to his eye.

But due to negative refraction, he will be able to see it since the refracted ray now reaches his eye.

Q. Is Snell’s law applicable for negative refraction ?

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 1
Snell’s law is satisfied for the materials having a negative refractive index, but the direction of the refracted light ray is ‘mirror-imaged’ about the normal to the surface.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 2

Read the following text and answer the following questions on the basis of the same:

Negative Refractive Index: One of the most fundamental phenomena in optics is refraction. When a beam of light crosses the interface between two different materials, its path is altered depending on the difference in the refractive indices of the materials. The greater the difference, the greater the refraction of the beam. For all known naturally occurring materials the refractive index assumes only positive values. But does this have to be the case?

In 1967, Soviet physicist Victor Veselago hypothesized that a material with a negative refractive index could exist without violating any of the laws of physics.

Veselago predicted that this remarkable material would exhibit a wide variety of new optical phenomena. However, until recently no one had found such a material and Veselago’s ideas had remained untested. Recently, meta-material samples are being tested for negative refractive index. But the experiments show significant losses and this could be an intrinsic property of negativeindex materials.

Snell’s law is satisfied for the materials having a negative refractive index, but the direction of the refracted light ray is ‘mirror-imaged’ about the normal to the surface.

There will be an interesting difference in image formation if a vessel is filled with “negative water” having refractive index – 1.33 instead of regular water having refractive index 1.33.

Say, there is a fish in a vessel filled with negative water. The position of the fish is such that the observer cannot see it due to normal refraction since the refracted ray does not reach to his eye.

But due to negative refraction, he will be able to see it since the refracted ray now reaches his eye.

Q. When the angle of incidence will be equal to angle of refraction for material having negative refraction index?

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 2
Like normal refraction, for material having negative refraction index also when the angle of incidence is equal to 0°, then angle of refraction will be equal to angle of incidence i.e. 0°.
1 Crore+ students have signed up on EduRev. Have you? Download the App
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 3

Read the following text and answer the following questions on the basis of the same:

Negative Refractive Index: One of the most fundamental phenomena in optics is refraction. When a beam of light crosses the interface between two different materials, its path is altered depending on the difference in the refractive indices of the materials. The greater the difference, the greater the refraction of the beam. For all known naturally occurring materials the refractive index assumes only positive values. But does this have to be the case?

In 1967, Soviet physicist Victor Veselago hypothesized that a material with a negative refractive index could exist without violating any of the laws of physics.

Veselago predicted that this remarkable material would exhibit a wide variety of new optical phenomena. However, until recently no one had found such a material and Veselago’s ideas had remained untested. Recently, meta-material samples are being tested for negative refractive index. But the experiments show significant losses and this could be an intrinsic property of negativeindex materials.

Snell’s law is satisfied for the materials having a negative refractive index, but the direction of the refracted light ray is ‘mirror-imaged’ about the normal to the surface.

There will be an interesting difference in image formation if a vessel is filled with “negative water” having refractive index – 1.33 instead of regular water having refractive index 1.33.

Say, there is a fish in a vessel filled with negative water. The position of the fish is such that the observer cannot see it due to normal refraction since the refracted ray does not reach to his eye.

But due to negative refraction, he will be able to see it since the refracted ray now reaches his eye.

Q. Who hypothesized that a material may have a negative refractive index ?

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 3
In 1967, Soviet physicist Victor Veselago hypothesized that a material with a negative refractive index could exist without violating any of the laws of physics.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 4

Read the following text and answer the following questions on the basis of the same:

Negative Refractive Index: One of the most fundamental phenomena in optics is refraction. When a beam of light crosses the interface between two different materials, its path is altered depending on the difference in the refractive indices of the materials. The greater the difference, the greater the refraction of the beam. For all known naturally occurring materials the refractive index assumes only positive values. But does this have to be the case?

In 1967, Soviet physicist Victor Veselago hypothesized that a material with a negative refractive index could exist without violating any of the laws of physics.

Veselago predicted that this remarkable material would exhibit a wide variety of new optical phenomena. However, until recently no one had found such a material and Veselago’s ideas had remained untested. Recently, meta-material samples are being tested for negative refractive index. But the experiments show significant losses and this could be an intrinsic property of negativeindex materials.

Snell’s law is satisfied for the materials having a negative refractive index, but the direction of the refracted light ray is ‘mirror-imaged’ about the normal to the surface.

There will be an interesting difference in image formation if a vessel is filled with “negative water” having refractive index – 1.33 instead of regular water having refractive index 1.33.

Say, there is a fish in a vessel filled with negative water. The position of the fish is such that the observer cannot see it due to normal refraction since the refracted ray does not reach to his eye.

But due to negative refraction, he will be able to see it since the refracted ray now reaches his eye.

Q. A ray incident on normal glass and “negative glass” at an angle 60°. If the magnitude of angle of refraction in normal glass is 45° then, what will be the magnitude of angle of refraction in the “negative glass”?

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 4
The magnitude of angle of refraction in normal “negative glass” will also be 45°, but the direction of the refracted light ray is ‘mirror-imaged’ about the normal to the surface.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 5

Read the following text and answer the following questions on the basis of the same:

Negative Refractive Index: One of the most fundamental phenomena in optics is refraction. When a beam of light crosses the interface between two different materials, its path is altered depending on the difference in the refractive indices of the materials. The greater the difference, the greater the refraction of the beam. For all known naturally occurring materials the refractive index assumes only positive values. But does this have to be the case?

In 1967, Soviet physicist Victor Veselago hypothesized that a material with a negative refractive index could exist without violating any of the laws of physics.

Veselago predicted that this remarkable material would exhibit a wide variety of new optical phenomena. However, until recently no one had found such a material and Veselago’s ideas had remained untested. Recently, meta-material samples are being tested for negative refractive index. But the experiments show significant losses and this could be an intrinsic property of negativeindex materials.

Snell’s law is satisfied for the materials having a negative refractive index, but the direction of the refracted light ray is ‘mirror-imaged’ about the normal to the surface.

There will be an interesting difference in image formation if a vessel is filled with “negative water” having refractive index – 1.33 instead of regular water having refractive index 1.33.

Say, there is a fish in a vessel filled with negative water. The position of the fish is such that the observer cannot see it due to normal refraction since the refracted ray does not reach to his eye.

But due to negative refraction, he will be able to see it since the refracted ray now reaches his eye.

Q. Which of the following is the intrinsic property of negative-index materials?

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 5
Recently, meta-material samples are being tested for negative refractive index. The experiments show significant losses and this is an intrinsic property of negative index materials.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 6

Read the following text and answer the following questions on the basis of the same:

First Surface Mirror:

Normally we use back surface mirrors. These are considered low precision mirrors because they actually have two reflecting surfaces. The first reflecting surface is the initial surface on the pane of glass where a small percentage of light is reflected off the surface. The second reflecting surface is the aluminium coating where a high percentage of light is reflected off the surface.

This dual reflection effect of a low precision mirror causes a loss of contrast and image distortion that is undesirable in high precision applications like rear projection systems, scanners and reflecting telescopes. In these cases good image quality is highly preferred, and this is where a front surface mirror is desired for clarity and single image reflection. First surface mirrors are quite common in professional optics. However, compared with back surface mirrors, they have the important disadvantage of being substantially more sensitive. The front surface may be touched, and a metal coating on the front surface is substantially more sensitive than a bare glass surface. For example, fingerprints can easily cause oxidation of the metal.. Also, moisture or aggressive gases may cause oxidation of the mirror coating.

Q. Light incident on back surface mirror suffers:

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 6
This dual reflection and refraction effect of a low precision mirror causes a loss of contrast and image distortion.

Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 7

Read the following text and answer the following questions on the basis of the same:

First Surface Mirror:

Normally we use back surface mirrors. These are considered low precision mirrors because they actually have two reflecting surfaces. The first reflecting surface is the initial surface on the pane of glass where a small percentage of light is reflected off the surface. The second reflecting surface is the aluminium coating where a high percentage of light is reflected off the surface.

This dual reflection effect of a low precision mirror causes a loss of contrast and image distortion that is undesirable in high precision applications like rear projection systems, scanners and reflecting telescopes. In these cases good image quality is highly preferred, and this is where a front surface mirror is desired for clarity and single image reflection. First surface mirrors are quite common in professional optics. However, compared with back surface mirrors, they have the important disadvantage of being substantially more sensitive. The front surface may be touched, and a metal coating on the front surface is substantially more sensitive than a bare glass surface. For example, fingerprints can easily cause oxidation of the metal.. Also, moisture or aggressive gases may cause oxidation of the mirror coating.

Q. Image formed of front coated mirror:

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 7
Front surface mirror produces bright, distinct and distortion less image. No ghost image is formed.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 8

Read the following text and answer the following questions on the basis of the same:

First Surface Mirror:

Normally we use back surface mirrors. These are considered low precision mirrors because they actually have two reflecting surfaces. The first reflecting surface is the initial surface on the pane of glass where a small percentage of light is reflected off the surface. The second reflecting surface is the aluminium coating where a high percentage of light is reflected off the surface.

This dual reflection effect of a low precision mirror causes a loss of contrast and image distortion that is undesirable in high precision applications like rear projection systems, scanners and reflecting telescopes. In these cases good image quality is highly preferred, and this is where a front surface mirror is desired for clarity and single image reflection. First surface mirrors are quite common in professional optics. However, compared with back surface mirrors, they have the important disadvantage of being substantially more sensitive. The front surface may be touched, and a metal coating on the front surface is substantially more sensitive than a bare glass surface. For example, fingerprints can easily cause oxidation of the metal.. Also, moisture or aggressive gases may cause oxidation of the mirror coating.

Q. Precision of back surface mirrors is:

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 8
Normally, we use back surface mirrors. These are considered low precision mirrors because they actually have two reflecting surfaces.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 9

Read the following text and answer the following questions on the basis of the same:

First Surface Mirror:

Normally we use back surface mirrors. These are considered low precision mirrors because they actually have two reflecting surfaces. The first reflecting surface is the initial surface on the pane of glass where a small percentage of light is reflected off the surface. The second reflecting surface is the aluminium coating where a high percentage of light is reflected off the surface.

This dual reflection effect of a low precision mirror causes a loss of contrast and image distortion that is undesirable in high precision applications like rear projection systems, scanners and reflecting telescopes. In these cases good image quality is highly preferred, and this is where a front surface mirror is desired for clarity and single image reflection. First surface mirrors are quite common in professional optics. However, compared with back surface mirrors, they have the important disadvantage of being substantially more sensitive. The front surface may be touched, and a metal coating on the front surface is substantially more sensitive than a bare glass surface. For example, fingerprints can easily cause oxidation of the metal.. Also, moisture or aggressive gases may cause oxidation of the mirror coating.

Q. In professional optics:

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 9
First surface mirrors are quite common in professional optics.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 10

Read the following text and answer the following questions on the basis of the same:

First Surface Mirror:

Normally we use back surface mirrors. These are considered low precision mirrors because they actually have two reflecting surfaces. The first reflecting surface is the initial surface on the pane of glass where a small percentage of light is reflected off the surface. The second reflecting surface is the aluminium coating where a high percentage of light is reflected off the surface.

This dual reflection effect of a low precision mirror causes a loss of contrast and image distortion that is undesirable in high precision applications like rear projection systems, scanners and reflecting telescopes. In these cases good image quality is highly preferred, and this is where a front surface mirror is desired for clarity and single image reflection. First surface mirrors are quite common in professional optics. However, compared with back surface mirrors, they have the important disadvantage of being substantially more sensitive. The front surface may be touched, and a metal coating on the front surface is substantially more sensitive than a bare glass surface. For example, fingerprints can easily cause oxidation of the metal.. Also, moisture or aggressive gases may cause oxidation of the mirror coating.

Q. The front surface coating:

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 10
The front surface mirror is substantially more sensitive than a bare glass surface. For example, fingerprints can easily cause oxidation of the metal. Also, moisture or aggressive gases may cause oxidation of the mirror coating.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 11

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.

Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.

Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 11
Focal length of the objective lens, f1 = 2.0 cm

Focal length of the eyepiece, f2 = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

(a) Least distance of distinct vision, d' = 25

∴ Image distance for the eyepiece, v2 = −25 cm

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

∴ u2 = −5 cm

Image distance for the objective lens, v1 = d + u2 = 15 − 5 = 10 cm

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

∴ u1 = −2.5 cm

Magnitude of the object distance, |u1| = 2.5 cm

Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 12

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.

Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.

Q. The intermediate image formed by the objective of a compound microscope is

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 12
The intermediate image formed by the objective of a compound microscope is real, inverted and magnified.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 13

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.

Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.

Q. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will be

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 13
Focal length of the objective lens, f1 = 2.0 cm

Focal length of the eyepiece, f2 = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

(a) Least distance of distinct vision, d' = 25

∴ Image distance for the eyepiece, v2 = −25 cm

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

∴ u2 = −5 cm

Image distance for the objective lens, v1 = d + u2 = 15 − 5 = 10 cm

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

∴ u1 = −2.5 cm

Magnitude of the object distance, |u1| = 2.5 cm

The magnifying power of a compound microscope is given by the relation:

= 4 × (1 + 4)

= 20

Hence, the magnifying power of the microscope is 20.

(b) The final image is formed at infinity.

∴ Image distance for the eyepiece, v2 = ∞

Object distance for the eyepiece = u2

According to the lens formula, we have the relation:

∴ u2 = −6.25 cm

Image distance for the objective lens, v1 = d + u2 = 15 − 6.25 = 8.75 cm

Object distance for the objective lens = u1

According to the lens formula, we have the relation:

Magnitude of the object distance, |u1| = 2.59 cm

Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 14

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.

Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.

Q. What is the magnifying power of the microscope in case of least distinct vision?

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 14
The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14cm. If the least distance of distinct vision is 20cm.
Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 15

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.

Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.

Q. The magnifying power of a compound microscope increases with

Detailed Solution for Case Based Questions Test: Ray Optics & Optical Instruments - 2 - Question 15
We know the formula for magnification of compound microscope,

Here, l is the length of the tube, fo is the focal length of the objective lens, D is the distance of distinct vision and fe is the focal length of the eyepiece lens.

We can write the proportionality of the terms in the above formula as,

From the above formula, we can conclude that the magnifying power of the compound microscope increases when the focal lengths of both objective and eyepiece lenses decrease.

So, the correct answer is option (D).

9 docs|1272 tests
Information about Case Based Questions Test: Ray Optics & Optical Instruments - 2 Page
In this test you can find the Exam questions for Case Based Questions Test: Ray Optics & Optical Instruments - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Case Based Questions Test: Ray Optics & Optical Instruments - 2, EduRev gives you an ample number of Online tests for practice

Top Courses for NEET

Download as PDF

Top Courses for NEET