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Physics M.C.Q (1 Marks)

Ray Optics and Optical Instruments · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 176 questions.

Questions · Page 2 of 4

M.C.Q (1 Marks)

MCQ 511 Mark

Which of the following quantities increase when wavelength is increased? Consider only the magnitudes.

  • A
    The power of a converging lens.
  • B
    The focal length of a converging lens.
  • C
    The power of a diverging lens.
  • D
    The focal length of a diverging lens.
Answer
  1. The focal length of a converging lens.
  1. The focal length of a diverging lens.

Explanation:

The focal length of a lens is inversely proportional to the refractive index of the lens and the refractive index of the lens is inversely proportional to the square of wavelength. Therefore, the focal length is directly dependent on wavelength; it increases when the wavelength is increased.

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MCQ 521 Mark

A magnifying glass is used to read the newspaper. As it is moved far away from the newspaper:

  • A
    Text becomes blurred and magnification reduces.
  • B
    Text becomes more focussed and magnification reduces.
  • C
    Text becomes more focused and magnification increases.
  • D
    Text becomes blurred and magnification increases.
Answer
  1. Text becomes blurred and magnification reduces.

Explanation:

As the magnifying lens(convex lens) is moved far away from the eye, the image formed is real and inverted and is formed inside the human eye and hence it blurs. Since the object is far away from the mirror, the image formed is diminished and hence magnification reduces.

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MCQ 531 Mark

The instrument that is based on the principle that when an object is placed between first principal focus and the optic centre of convex lens, an upright, virtual and enlarged image on the same side of the object is formed, is:

  • A
    Telescope
  • B
    Projector
  • C
    Camera
  • D
    Simple microscope
Answer
  1. Simple microscope

Explanation:

In astronomical telescope 2 convex lens called eyepiece and objective lens are used and object is placed before eyepiece lens, such that final image inverted, a camera and eye also form inverted image on the screen.

Whereas simple microscope gives an erect, virtual and enlarged image of the object placed between first principal focus and the optic nerve of the convex lens.

In a projector, the image formed is real, inverted magnified on the other side of the lens. This inverted image is again inverted by the film.

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MCQ 541 Mark

For reading small letters with a lens:

  • A
    One has to keep a convex lens at a distance between F and 2F from the book.
  • B
    One has to keep a concave lens at a distance less than the focal length from the book
  • C
    One has to keep concave lens at a distance between F and 2F from the book
  • D
    One has to keep a convex lens at a distance less than the an focal length from the book
Answer
  1. One has to keep a convex lens at a distance less than the an focal length from the book

Explanation:

A concave lens always produces virtual, erect and diminished images and the decrease in the size of the image depends on the position of the object.

Concave lens will shrink the size of the already small letters.

A convex lens produces real and virtual, erect and inverted, diminished, same sized and magnified image of the object, depending upon the position of the object on the principal axis.

When the object is placed between F and 2F of convex lens, an enlarged but inverted image of the object is formed. The magnified image makes it easier to read small letters but the inverted image is undesirable.

When the object is placed at a distance less than the focal length of convex lens, an enlarged and erect image of the object is formed, which makes it easier to read small letters.

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MCQ 551 Mark

One cannot see through fog because .......... :

  • A
    Fog absorbs light.
  • B
    Refractive index of fog is unity.
  • C
    Light suffers total internal reflection at the droplets in fog.
  • D
    Light is scattered by the droplets in fog.
Answer
  1. Light is scattered by the droplets in fog.

Explanation:

We cannot see through fog because of scattering.

Atoms and molecules in the air, including anything carried in the air like dust or smoke, will scatter light. Water droplets, as they are present in fog, also scatter light.

The light falling on an object and reflected to a viewer can be scattered to heck and back before it gets to the place where it can be 'seen' by an observer.

So the observer just sees a 'whiteout' instead of being able to make out anything beyond a few meters or so.

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MCQ 561 Mark

The phenomenon of light bending due to change of medium is called:

  • A
    Reflection
  • B
    Refraction
  • C
    Dispersion
  • D
    Total internal reflection
Answer
  1. Refraction

Explanation:

When light enters from one medium to another, its speed and direction changes, and hence the light seems to be bending towards or away from the normal. This phenomenon is called refraction of light.

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MCQ 571 Mark

When a wave travels from one medium to another, the quantity which will not change is its:

  • A
    Amplitude
  • B
    Velocity
  • C
    Frequency
  • D
    Intensity
Answer
  1. Frequency

Explanation:

When a wave travels from one medium to another, the amplitude, wavelength, velocity and intensity change but frequency does not change.

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MCQ 581 Mark

You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90º. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?

  • A
    The beam of red light would undergo total internal reflection.
  • B
    The beam of red light would bend towards normal while it gets refracted through the second medium.
  • C
    The beam of blue light would undergo total internal reflection.
  • D
    The beam of green light would bend away from the normal as it gets refracted through the second medium.
Answer
  1. The beam of blue light would undergo total internal reflection.

Solution:

Key Concept: According to Couchy relationship,

$\lambda\propto\frac{1}{\mu}$

Smaller the wavelengh higher the refractive index and consequently smaller the critical angle.

We know $\text{v}=\text{f}\lambda$ the frequency of wave remains unchanged with medium hence $\text{v}\propto\lambda$.

The critical angle, sin $\text{C}=\frac{1}{\mu}$

Also, velocity of light, $\text{v}\propto\frac{1}{\mu}$

According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength. As $\lambda_{\text{blue}}<\lambda_\text{yellow}$ hence $\text{v}_{\text{blue}}<\text{v}_\text{yellow}$, it means $\mu_{\text{blue}}<\mu_\text{yellow}$.

It means critical angle for blue is less than yellow colour, the critical angle is least which facilitates total internal reflection for the beam of blue light.

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MCQ 591 Mark

Which of the following has a larger chromatic aberration?

  • A
    Crown-glass lens
  • B
    Flint-glass lens
  • C
    Both have equal chromatic aberration
  • D
    Insufficient information
Answer
  1. Flint-glass lens

Explanation:

Flint-glass lens has a larger chromatic aberration because the dispersive power of flint-glass is higher.

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MCQ 601 Mark

The focal length of a lens of refractive index $\mu$ is f. If the lens is immersed in a liquid of refractive index $\mu$ then the focal length of lens is:

  • A
    $\frac{\text{f}}{2}$
  • B
    2f
  • C
    Infinite
  • D
    zero
Answer
  1. Infinite

Explanation:

$\frac{1}{\text{f}}​=(\text{n}−1)\Big( \frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$

where, n is refractive index of lens material with respect to its surrounding medium.

If lens is immersed in the liquid of refractive index equal to its own then n = 1 and hence above equation becomes: $\frac{1}{\text{f}}=0$

⇒ f = $\infty$ 

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MCQ 611 Mark

Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?

  • A
    $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{uf}}$
  • B
    $\frac{1}{\text{v}^2}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
  • C
    $\frac{1}{\text{v}-\text{t}}-\frac{1}{\text{u}+\text{t}}=\frac{1}{\text{f}}$
  • D
    $\frac{1}{\text{v}}-\frac{1}{\text{u}}+\frac{\text{t}}{\text{uv}}=\frac{\text{t}}{\text{f}}$
Answer
  1. $\frac{1}{\text{v}-\text{t}}-\frac{1}{\text{u}+\text{t}}=\frac{1}{\text{f}}$

Explanation:

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MCQ 621 Mark

The working of optical instruments like camera, microscope, telescope, etc. having glass lenses is based on a phenomenon of light. Identify the phenomenon.

  • A
    Reflection
  • B
    Refraction
  • C
    Dispersion
  • D
    Scattering
Answer
  1. Reflection
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MCQ 631 Mark

In the white light of sun, maximum scattering by the air molecules present in the earth's atmosphere is for:

  • A
    Red colour
  • B
    Yellow colour
  • C
    Green colour
  • D
    Blue colour
Answer
  1. Blue colour

Explanation

By Rayleigh's Criterion intensity of scattered light is given by, $\text{I}\propto\frac{1}{\lambda^4}​$

Intensity of scattered light is more for smaller wavelengths of light.

In the visible spectrum, the blue side of the spectrum has a smaller wavelength compared to the red side. Hence blue light is scattered more by air molecules present in earth's atmosphere.

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MCQ 641 Mark

Figure shows the ray diagram of a:

  • A
    Simple microscope
  • B
    Compound microscope
  • C
    Simple Telescope
  • D
    Compound Telescope
Answer
  1. Simple microscope

Explanation:

The given figure is representing the ray diagram of a simple microscope.

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MCQ 651 Mark

The relation among u, v and f for a mirror is:

  • A
    $\text{f}=\frac{\text{uv}}{\text{u+v}}$
  • B
    $\text{v}=\frac{\text{fu}}{\text{u+f}}$
  • C
    $\text{u}=\frac{\text{fv}}{\text{f+v}}$
  • D
    All of these
Answer
  1. $\text{f}=\frac{\text{uv}}{\text{u+v}}$

Explanation:

As we know,

$\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{u}}$

$\frac{1}{\text{f}}=\frac{\text{u+v}}{\text{uv}}$

$\text{f}=\frac{\text{uv}}{\text{u+v}}$

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MCQ 661 Mark

A point source of light is placed in front of a plane mirror:

  • A
    All the reflected rays meet at a point when produced backward.
  • B
    Only the reflected rays close to the normal meet at a point when produced backward.
  • C
    Only the reflected rays making a small angle with the mirror, meet at a point when produced backward.
  • D
    Light of different colours make different images.
Answer
  1. All the reflected rays meet at a point when produced backward.

Explanation:

​​​​​​​

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MCQ 671 Mark

The maximum focal length of the eye-lens of a person is greater than its distance from the retina. The eye is:

  • A
    Always strained in looking at an object.
  • B
    Strained for objects at large distances only.
  • C
    Strained for objects at short distances only.
  • D
    Unstrained for all distances.
Answer
  1. Aways strained in looking at an object.

Explanation:

The maximum focal length of a normal eye is equal to the distance of the lens from the retina. In case it is greater than the distance, the eye will be strained while focusing the objects on the retina that is at a fixed distance from the eye lens.

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MCQ 681 Mark

A convex lens is made of a material having refractive index 1.2. Both the surfaces of the lens are convex. If it is dipped into water $(\mu=1.33),$ it will behave like:

  • A
    A convergent lens.
  • B
    A divergent lens.
  • C
    A rectangular slab.
  • D
    A prism.
Answer
  1. a divergent lens.

Explanation:

Here P, P1 & P2 are the Power of Lenses.

P = P1 + P2

$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}=\frac{1}{\text{f}_2}$

$(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$

$(1.2-1)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$

$\frac{1}{\text{f}}=\frac{2}{5\text{R}}-\frac{1}{3\text{R}}$

$\frac{1}{\text{f}}=\frac{6-5}{15\text{R}}$

$\text{f}=15\text{R}$

Focal lenght of combined is positive, but it's magnitude in capair to f1 & f2 is High. So it will be hare like a divergent lens.

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MCQ 691 Mark

When a ray of light passes from a denser to a rarer medium:

  • A
    It bends away from the normal.
  • B
    It bends away from incident ray.
  • C
    It bends toward the incident ray.
  • D
    Goes parallel to the interface separating two media.
Answer
  1. It bends away from the normal.

Explanation:

When a ray of light passes from a denser to a rarer medium, some part of it gets refracted into the rarer medium such that it bends away from the normal. Some part of it gets reflected back into the denser medium. The light reflected back into the denser medium is said to be internally reflected.

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MCQ 701 Mark

The perpendicular distance between the original path of the incident ray and the convergent ray of light coming out of a glass slab is called:

  • A
    Refraction
  • B
    Lateral displacement
  • C
    Total internal reflection
  • D
    None
Answer
  1. Lateral displacement

Explanation:

The perpendicular distance between the original path of the incident ray and the emergent ray of light coming out of a glass slab is called lateral displacement. It is proportional to the thickness glass slab.

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MCQ 711 Mark

An object approaches a convergent lens from the left of the lens with a uniform speed 5m/s and stops at the focus. The image.

  • A
    Moves away from the lens with an uniform speed 5m/s.
  • B
    Moves away from the lens with an uniform accleration.
  • C
    Moves away from the lens with a non-uniform acceleration.
  • D
    Moves towards the lens with a non-uniform acceleration.
Answer
  1. Moves away from the lens with a non-uniform acceleration.

Solution:

If an object approaches a convergent lens from the left of the lens with a uniform speed of 5m/s, then the image moves away from lens with a non-uniform acceleration.

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MCQ 721 Mark

A person using a lens as a simple microscope sees an:

  • A
    Inverted virtual image
  • B
    Inverted real magnified image
  • C
    Upright virtual image
  • D
    Upright real magnified image
Answer
  1. Upright virtual image

Explanation:

A simple microscope is just a convex lens with object lying between optical centre and focus of the lens.

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MCQ 731 Mark

A convex lens of ______ focal length gives a greater magnification than lenses of _______ focal length.

  • A
    Short, short
  • B
    Short, long
  • C
    Long, short
  • D
    Long, long
Answer
  1. Short, long

Explanation:

A convex lens of short focal length gives a greater magnification than lenses of long focal length.

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MCQ 741 Mark

Which of the following quantities related to a lens depend on the wavelength or wavelengths of the incident light?

  • A
    Power.
  • B
    Focal length.
  • C
    Chromatic aberration.
  • D
    Radii of curvature.
Answer
  1. Power.
  2. Focal length.
  3. Chromatic aberration.

Explanation:

The focal length, power and chromatic aberration are dependent on the refractive index of the lens, which itself is dependent on the wavelength of the light.

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MCQ 751 Mark

A person A can clearly see objects between 25cm and 200cm. Which of the following may represent the range of clear vision for a person B having muscles stronger than A, but all other parameters of eye identical to that of A?

  • A
    25cm to 200cm.
  • B
    18cm to 200cm.
  • C
    25cm to 300cm.
  • D
    18cm to 300cm.
Answer
  1. 18cm to 200cm.

Explanation:

Person B has stronger ciliary muscles than person A. So, the muscles in his case can be strained more and the focal length of his eye can be reduced more compared to those of person A. While seeing far objects, the muscles are relaxed, so their strength will not affect the far point of the eye.

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MCQ 761 Mark

An ellipsometer is an instrument for.

  • A
    Measuring stellar distance.
  • B
    Measuring the path of celestial bodies.
  • C
    Ending curvature of elliptical surfaces.
  • D
    Studying thin films on a solid surface.
Answer
  1. Studying thin films on a solid surface.

Explanation:

An ellipsometer is a device used for studying thin films on a solid surface.

Ellipsometry is an optical technique for investigating the di electric properties of thin films.

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MCQ 771 Mark

In case of a real and inverted image, the magnification created by the mirror is:

  • A
    Positive
  • B
    Negative
  • C
    Unity
  • D
    Infinity
Answer
  1. Negative

Explanation:

Magnification of image created by mirror is defined as

$\text{m}=\frac{\text{size of object}}{\text{size of image}}$​

and in case of inverted image. Size of image is negative whereas size of object is positive. Hence , magnification produced is negative and it can be unity when object is placed at center of curvature and infinity when object is at focus.

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MCQ 781 Mark

Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because:

  • A
    The apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
  • B
    The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
  • C
    Some of the points of the object far away from the edge may not be visible because of total internal reflection.
  • D
    Water in a trough acts as a lens and magnifies the object.
Answer
  1. The apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
  2. The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
  3. Some of the points of the object far away from the edge may not be visible because of total internal reflection.

Solution:

Key concept: The light from the pencil is refracted when it passes from the water into air, bending away from the normal as it moves from high to low refractive index.

When light from the submerged object before reaching to the observer gets, refracted from water surface, the rays bend away from normal and the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air. Also the apparent depth of the .points close to the edge are nearer the surface of the water compared to the points away from the edge.

As we move towards right, the angle of incident increases and becomes equal to critical angle. Hence some of the points of the object far away from the edge may not be visible because of total internal reflection.

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MCQ 791 Mark

The phenomenon due to which a ray of light .......... from its path while travelling from one optical medium to another optical medium is called refraction.

  • A
    Reflected
  • B
    Deviates
  • C
    Normally
  • D
    Less
Answer
  1. Deviates

Explanation:

Refraction is a phenomenon in which when a ray passes from one medium to another it bends away from its straight-line path due to the difference in optical densities or refractive indices of the two mediums. This bending is known as deviation.

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MCQ 801 Mark

The lateral displacement depends on ______:

  • A
    Thickness of the medium
  • B
    Refractive index of the medium
  • C
    Angle of incidence
  • D
    All the above
Answer
  1. All the above

Explanation:

Lateral displacement due to a slab is given by $\delta=\text{t}(\mu−1)$; where t is thickness of the medium and μ is refractive index of medium.

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MCQ 811 Mark

prism can produce a minimum deviation $\delta$ in a light beam. If three such prisms are combined, the minimum deviation that can be produced in this beam is:

  • A
    $0$
  • B
    $\delta$
  • C
    $2\delta$
  • D
    $3\delta$
Answer
  1. $\delta$

Explanation:

In combination (refractive angles of prisms reversed with respect to each other), the deviations through two prisms cancel out each other and the net deviation is due to the third prism only.

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MCQ 821 Mark

The phenomenon of change in the _____of light when it passes from one transparent medium to another is refraction.

  • A
    Speed
  • B
    Direction
  • C
    Both A & B
  • D
    None of the above
Answer
  1. Both A & B

Explanation:

The phenomenon of change in the path of light when it passes from one transparent medium to another is refraction. This change in path is due to the change in the speed of light in different media.

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MCQ 831 Mark

If a ray of light goes from a rarer medium to a denser medium, will it bend towards the normal or away from it?

  • A
    Bends away from the normal
  • B
    Bends towards the normal
  • C
    Goes undeviated
  • D
    Is reflected back
Answer
  1. Bends towards the normal

Explanation:

Refraction is the bending of light rays after entering a medium where its speed is different. Due to refraction of light, when a ray of light passes from a rarer medium to a denser medium, bends towards the normal to the boundary between the two media. The amount of bending depends on the indices of refraction of the two media. Hence, when a ray of light from air enters a denser medium, it bends towards the normal.

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MCQ 841 Mark

A symmetric double convex lens is cut in two equal parts by a plane containing the principal axis. If the power of the original lens was 4D, the power of a divided lens will be:

  • A
    2D
  • B
    3D
  • C
    4D
  • D
    5D.
Answer
  1. 4D

Explanation:

Before cut

$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{2}{\text{R}}\Big)=4\text{D}$

After cut

$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_1$

$\&\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_2$

Power of a divided lens will be = P1 + P2

$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)$

$=4\text{D}$

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MCQ 851 Mark

Mark the correct options:

  • A
    If the far point goes ahead, the power of the divergent lens should be reduced.
  • B
    If the near point goes ahead, the power of the convergent lens should be reduced.
  • C
    If the far point is 1 m away from the eye, divergent lens should be used.
  • D
    If the near point is 1 m away from the eye, divergent lens should be used.
Answer

  1. If the far point goes ahead, the power of the divergent lens should be reduced.

  1. If the far point is 1m away from the eye, the divergent lens should be used.

Explanation:

As the far point (x) is shifited ahead, the focal length (f) will be increased.

thus. we have

$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$

$\Rightarrow \frac{1}{\text{f}}=\frac{1}{-\text{x}}-\frac{1}{-\infty}$

$\Rightarrow \frac{1}{\text{f}}=\frac{1}{-\text{x}}$

$\Rightarrow\text{f}=-\text{x}$

As power (p) is equal to reciprocal of the focal length, it will be reduced, Also, because the focal length is negative, the lens used will be divergent when the far point is 1m away.

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MCQ 861 Mark

In case of a virtual and erect image, the magnification created by the mirror is:

  • A
    Positive
  • B
    Negative
  • C
    Unity
  • D
    Infinity
Answer
  1. Positive

Explanation:

Magnifaction of a mirror is defined as

$\text{m}=\frac{\text{size of image​}}{\text{size of object}}$

and since, in case of virtual and errect image, size of image and object both are positive. Hence magnification created by mirror is positive.

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MCQ 871 Mark

A screen is placed a distance 40cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens:

  • A
    Must be less than 10cm.
  • B
    Must be greater than 20cm.
  • C
    Must not be greater than 20cm.
  • D
    Must not be less than 10cm.
Answer
  1. Must be greater than 20cm.

Explanation:

$\text{v}=(40-4)$

$\frac{1}{\text{f}}=\frac{1}{40-4}-\frac{1}{(-\text{u})}$

$\frac{\text{df}}{\text{du}}=0$ for f minimum.

$\frac{\text{df}}{\text{du}}=1-\frac{\text{u}}{20}=0$

$\text{u}=20$

$\text{f}_{\text{min}}=10\text{cm}$

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MCQ 881 Mark

The danger signals installed at the top of tall buildings are red in colour. These can be easily seen from a distance because among all other colours, the red light:

  • A
    Is scattered the most by smoke or fog.
  • B
    Is scattered the least by smoke or fog.
  • C
    Is absorbed the most by smoke or fog.
  • D
    Moves fastest in air.
Answer
  1. Is scattered the least by smoke or fog.

Explanation:

The amount of scattering is inversely proportional to the fourth power of wavelength. And since the wavelength of red has the longest wavelength, the amount of scattering becomes smaller.

Danger signals installed at the top of tall buildings or traffic signals are red so that they can be easily seen from a distance.

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MCQ 891 Mark

Mark the correct options:

  • A
    If the incident rays are converging, we have a real object.
  • B
    If the final rays are converging, we have a real image.
  • C
    The image of a virtual object is called a virtual image.
  • D
    If the image is virtual, the corresponding object is called a virtual object.
Answer
  1. If the final rays are converging, we have a real image.

Explanation:

This is because a real image is formed by converging reflected/ refracted rays from a mirror/ lens.

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MCQ 901 Mark

The angular dispersion produced by a prism:

  • A
    Increases if the average refractive index increases.
  • B
    Increases if the average refractive index decreases.
  • C
    Remains constant whether the average refractive index increases or decreases.
  • D
    Has no relation with average refractive index.
Answer
  1. Increases if the average refractive index increases.

Explanation:

If $\mu$ is the average refractive index and A is the angle of prism, then the angular dispersion produced by the prism is given by $\delta=(\mu-1)\text{A}.$

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MCQ 911 Mark

An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by:

  • A
    A screw gauge provided on the microscope
  • B
    A standard laboratory scale
  • C
    A vernier scale provided on the microscope
  • D
    A meter scale provided on the microscope
Answer
  1. A vernier scale provided on the microscope

Explanation:

The travelling microscope moves horizontally on a main scale provided with the vernier scale provided with the microscope that's why, In a travelling microscope to find the refractive index of glass we measure distance by a vernier scale provided on the microscope.

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MCQ 921 Mark

A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in;

  • A
    A larger angle to be subtended by the object at the eye and hence viewed in greater detail.
  • B
    The formation of a virtual erect image.
  • C
    Increase in the field of view.
  • D
    Infinite magnification at the near point.
Answer
  1. A larger angle to be subtended by the object at the eye and hence viewed in greater detail.
  2. The formation of a virtual erect image.

Solution:

Key concept: A magnifying glass is a single convex lens of lesser focal length.

For magnification when final image is formed at D and $\infty(\text{i.e., m}_\text{D}\text{ and m}_\infty)$.

$\text{m}_\text{D}=\Big(1+\frac{\text{D}}{\text{f}}\Big)_\text{max}\text{ and }\text{m}_\infty=\Big(\frac{\text{D}}{\text{f}}\Big)_\text{min}$

When a magnifying glass is used, the object to be viewed can be brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence, viewed in greater detail. Moreover, the formation of a virtual erect and enlarged image takes place.

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MCQ 931 Mark

The distance between the extreme points on the periphery of the mirror is called:

  • A
    Focal length
  • B
    Radius of curvature
  • C
    Principal section
  • D
    None of these
Answer
  1. Principal section

Explanation:

Principal section is also defined as the normal 'side view' of the mirror for a ray diagram. In the diagram AB is the principal section.

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MCQ 941 Mark

How can we explain the reddish appearance of sun at sunrise or sunset? 

  • A
    Scattering of blue light is more than the scattering of red light.
  • B
    Scattering of red light is more than the scattering of blue light.
  • C
    Intensity of sun reduces during sunrise and sunset.
  • D
    Due to the view angle, it appears blue.
Answer
  1. Scattering of blue light is more than the scattering of red light.

Explanation:

During sunrise or sunset, the light has to pass through greater distance in the atmosphere. The blue light is removed as it gets scattered the most while the red colour is less scattered and reaches the observer. Thus, we find reddish colour of the sun during sunrise or sunset.

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MCQ 951 Mark

Why does a straight rod appear bent in water?

  • A
    Due to reflection of light
  • B
    Due to refraction of light
  • C
    Due to variable refractive index of water.
  • D
    None of the above
Answer
  1. Due to refraction of light

Explanation:

This phenomenon occurs due to the property of light called refraction of light. When a stick is immersed in water, in actually we are putting it from rarer medium to denser medium. So, when the rays of light pass from a rarer medium to the denser medium they move towards the normal, the part of stick immersed in water appears to bend when immersed in water and this refraction causes an apparent shift in the position of the part of the rod within the water.

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MCQ 961 Mark

A plane mirror produces an image that is:

  • A
    Real, inverted and larger than the object.
  • B
    Real, upright and same size as the object.
  • C
    Real upright and smaller than the object.
  • D
    Virtual, upright and the same size of the object.
Answer
  1. Virtual, upright and the same size of the object.

Explanation:

 So a plane mirror always forms a virtual, upright and same sized image. ​

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MCQ 971 Mark

The rays parallel and close to the principal axis are called:

  • A
    Converging rays
  • B
    Diverging rays
  • C
    Coherent rays
  • D
    Paraxial rays
Answer
  1. Paraxial rays

Explanation:

The rays parallel and close to the principal axis are called paraxial rays. The rays parallel but not close to the principal axis are called peripheral rays.

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MCQ 981 Mark

An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2cm.

  • A
    The length of the telescope tube is 20.02m.
  • B
    The magnification is 1000.
  • C
    The image formed is inverted.
  • D
    An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image.
Answer
  1. The length of the telescope tube is 20.02m.
  2. The magnification is 1000.
  3. The image formed is inverted.

Solution:

Key concept:

  1. Used to see heavently bodies.

  2. fobjective > feye lens and dobjective > deye lens.

  3. Intermediate image is real, inverted and small.

  4. Final image is virtual, inverted and small.

  5. Magnification: $\text{m}_\text{D}=-\frac{\text{f}_0}{\text{f}_\text{e}}\Big(1+\frac{\text{f}_\text{e}}{\text{D}}\Big)$ and $\text{m}_\infty=-\frac{\text{f}_0}{\text{f}_\text{e}}$.

  6. Length: $\text{L}_\text{D}=\text{f}_0+\text{u}_\text{e}=\text{f}_0+\frac{\text{f}_\text{e}\text{D}}{\text{f}_\text{e}+\text{D}}$ and $\text{L}_\infty=\text{f}_0+\text{f}_\text{e}$.

The length of the telescope tube is f0 + fe = 20 + (0.02) = 20.02m

Also, $\text{m}=\frac{20}{0.02}=1000$

Also, the image formed is inverted.

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MCQ 991 Mark

When a ray of light passes from a denser to a rarer medium, some part of it gets ....... into the denser medium:

  • A
    Reflected
  • B
    Refracted
  • C
    Both
  • D
    None
Answer
  1. Reflected

Explanation:

When a ray of light passes from a denser to a rarer medium, some part of it gets refracted into the rarer medium such that it bends away from the normal. Some part of it gets reflected back into the denser medium. The light reflected back into the denser medium is said to be internally reflected.

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MCQ 1001 Mark

The focal length of the objective of a compound microscope if fo and its distance from the eyepiece is L. The object is placed at a distance u from the objective. For proper working of the instrument:

  • A
    L < u
  • B
    L > u
  • C
    fo < L < 2fo
  • D
    L > 2fo
Answer
  1. L > u
  1. L > 2fo

Explanation:

In a compound microscope, the objective lens of a short focal length, fo is used. The focal length of the objective lens is less than the focal length of the eyepiece, f​e,. The object is placed at a distance slightly greater than its focal length. The real, inverted image of the object forms somewhere in front of the eyepiece at a distance less than its focal length. This image acts as its object and the final image forms in between length, L of the microscope.

$\therefore$ L > fo​ + fe > 2fo​

Also, L > u

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