Ray Optics MCQ Quiz - Objective Question with Answer for Ray Optics - Download Free PDF

Calculation:

Objective focal length (f_o) = 2 cm

Eyepiece focal length (f_e) = 4 cm

Tube length (L) = 40 cm

Distance of distinct vision of eye (D) = 25 cm

Formula for magnification (m) of the microscope:

m = (L / f_o) × (D / f_e)

⇒ m = (40 / 2) × (25 / 4)

⇒ m = 125

Final Answer: The magnification in the microscope is 125.

Calculation: 

For a convex lens to always give a real image, the object must be placed beyond the focal point of the lens.

This is because a convex lens converges light rays to a point on the other side of the lens. When the object is placed beyond the focal point, the light rays after refraction converge to form a real image on the other side of the lens.

Let’s consider the options provided:

1. Optic centre
2. Focus
3. Radius of curvature
4. Centre of curvature

When the object is at the focus, the image formed is at infinity, and when the object is at the optic centre, the image formed is virtual and at the same position as the object.

For real images, the correct placement of the object is beyond the focus.

Therefore, the correct answer is option 2.

Concept Used:

Refraction and Reflection at the Interface:

n₁ sin(i) = n₂ sin(r)

tan(θₓ) = n₂ / n₁

When unpolarized light strikes a plane glass surface, the angle of incidence (i) and the angle of refraction (r) are related by Snell's law:

For the reflected and refracted rays to be perpendicular, the sum of the angle of reflection (r) and the angle of refraction (i) should be 90°.

This condition is known as the Brewster angle, which occurs when the refracted and reflected rays are perpendicular to each other.

The Brewster's angle (θₓ) can be given by the formula:

Where:

n₁ = Refractive index of air (approximately 1)

n₂ = Refractive index of the glass (typically 1.5 for glass)

SI Unit of angle: Degree (°)

Calculation:

Given,

n₁ = 1 (air)

n₂ = 1.5 (glass)

Using the formula for Brewster's angle:

tan(θₓ) = n₂ / n₁

⇒ tan(θₓ) = 1.5 / 1

⇒ tan(θₓ) = 1.5

Using a calculator, θₓ ≈ 56°

∴ The angle of incidence should be 56° for the reflected and refracted rays to be perpendicular.

Ans. (3)

Solution:

Concave mirror is used by a dentist such that f = -16 mm and u = -8 mm.

Using the mirror formula: 1/u + 1/v = 1/f

⇒ 1/v = 1/f - 1/u = 1/(-16) - (1/(-8))

⇒ 1/v = -1/16 + 1/8 = (-1 + 2)/16

⇒ 1/v = 1/16 ⇒ v = 16 mm

So, magnification m = -v/u = -16 / (-8) = 2

The correct answer is: Option 1) 27°

Concepts:

An astronomical telescope forms an image of a distant object by using an objective lens and an eyepiece lens. The angular magnification (M) of a telescope when the final image is formed at the least distance of distinct vision (D) can be given by:

M = (f_O / f_E) × (1 + f_E / D)

where f_O is the focal length of the objective and f_E is the focal length of the eyepiece. The angular size of the image is the product of the angular magnification and the angular size of the object.

Calculation:

Given:

f_O = 50 cm

f_E = 2 cm

D = 25 cm

Angular diameter of moon at the objective = 1/2°

First, calculate the angular magnification (M):

M = (50 / 2) × (1 + 2 / 25)

M = 25 × (1 + 0.08)

M = 25 × 1.08 = 27

Then, the angular size of the image = M × angular diameter of moon

Angular size of image = 27 × (1/2)°

Angular size of image = 13.5°

​The correct answer is Ciliary muscles.

Key Points

1. The cornea is a thin membrane through which the light rays enter our eyes. It forms the transparent bulge on the front surface of the eyeball.
2. Aqueous humor is the transparent watery fluid present between the cornea and eye lens. It is secreted from the ciliary muscles.
3. Ciliary muscles are smooth muscle fibers. When the ciliary muscles relax the lens becomes flat and when they contract the lens becomes thicker. This change in curvature of the eye lens changes the focal length of the eyes.
4. Iris is a muscle in our eye that controls the size of the pupil. It dilates and contracts because of which the size of the pupil varies.

I

• Power of accommodation: The ability of the eye lens to adjust its focal length is called its power of accommodation.
  • The ciliary muscles are capable of changing the curvature of the eye lens to some extent, they thus change the focal length of the eye lens.

EXPLANATION:

• The Ciliary muscles are responsible for adjusting the power of accommodation of the eye lens.

Additional Information

Different parts of the eyes and their functions are shown in the table:

The correct answer is It is the Centre of a hollow sphere of which the spherical mirror is a part.

Key Points

• The point in the center of the sphere from which the glass was sliced is known as the center of the curve and is denoted by the letter C in the illustration.
• This point is equidistant from all points on the reflecting face of the glass.
• The shaft illustration for the image conformation by a hollow glass when the size of an object and its image are the same is given above.
• Then, in the illustration, if we put the object in the center of the curve also the image will have the same size as that of the object.

Important Points

• Mirror Formula:

  • Mirror Formula is given by:-
    \(1/f=1/v+1/u\)1
    ​where,

    f = Focal length of the mirror.

    v = Distance of image from the mirror.

    u = Distance of the object from the mirror.

CONCEPT:

Refraction: 

• The phenomenon by which a ray of light bends its path when it travels from one transparent medium to another is called the refraction of light.
• The bending happens because the speed of light varies in different mediums.

Refractive index: 

• The ratio of the speed of light in the air by the speed of light in the medium is called the refractive index. More will be the ratio of refractive index between two mediums more will be the bending of light. 

Atmospheric Refraction: 

• The different layers of the atmosphere have a different refractive index.
•  The ray of light coming from some outer space changes continuously bends its path while passing through different atmospheric layers. 

EXPLANATION:

• The atmosphere of the earth is made of different layers.
• It is affected by winds, varying temperatures, and different densities as well.
• When light from a distant source (a star) passes through our turbulent (moving air) atmosphere, it undergoes refraction many times.
• When we finally perceive this light from a star, it appears to be twinkling.
• So, we can say that twinkling of stars is due to the atmospheric refraction of starlight.
• This is because some of the light rays reach us directly and some bend away from and toward us. It happens so fast that it gives a twinkling effect.​

Important Points 

• This twinkling effect does not happen for the sun or moon because these are very close to the earth and appear big.
• Stars are very far and appear to us as point size. So, the twinkling effect is visible.

The correct answer is is negative.Key Points

• The height of the image of an object below principal axis of a spherical mirror is negative.
• This is because the image formed by a spherical mirror below the principal axis is virtual and inverted.
• As a result, the height of the image will be negative.
• The position of the object will affect the location and size of the image, but it does not determine the height of the image.
• The height of the image will vary depending on where it is located with respect to the principal axis, and if the image is located below the principal axis, the height will be negative.

Additional Information

• The height of the image will vary depending on where it is located with respect to the principal axis.
• The position of the object will affect the location and size of the image, but it does not determine the height of the image.
• If the image is located above the principal axis, the height will be positive.
• Spherical mirrors are mirrors that have a curved surface, with the shape of a sphere.
• The principal axis is an imaginary line that passes through the center of the sphere and the vertex of the mirror.
• The height of the image refers to the distance between the top of the image and the principal axis, or between the bottom of the image and the principal axis.
• The height of the object refers to the distance between the top of the object and the principal axis, or between the bottom of the object and the principal axis.

The correct answer is Scattering.

Concept:

• The phenomenon of light associated with the appearance of the blue Colour of the sky is called scattering. 
• Scattering of light is the process of light scattered in a different direction according to its Wavelength and Frequency.
• As a result of Scattering, we see the Sun in Red Color during the sunset and sunrise and the sky in Blue Color.

Additional Information

 Interference: 

• ​It is the phenomenon that two waves interfere and superimpose each other.

Reflection:

• ​When light reflects from the surface at some angle is called reflection.

Refraction:

• ​It is the change in the direction of the wave when passing from one medium to another.

Given:

Position of Object for concave mirror \(u=-15cm\)

Radius of curvature,R\(=-20cm\)

Focal length \(F=\frac{-R}{2}=\frac{-20}{2}=-10cm\)

Size of Object, \(h_o\)=2cm

Concept:

The mirror formula is given by

• \(\frac{1}{F}=\frac{1}{u}+\frac{1}{v}\)

• \(\frac{h_i}{h_o}=-\frac{v}{u}\)

Explanation:

• \(\frac{1}{F}=\frac{1}{u}+\frac{1}{v}\implies\frac{1}{v}=\frac{1}{F}-\frac{1}{u}\)

• \(\frac{1}{v}=\frac{1}{-10}-\frac{1}{-15}=\frac{1}{-10}+\frac{1}{15}=\frac{1}{-30}\)

\(\implies v=-30cm\)

• \(\frac{h_i}{h_o}=-\frac{v}{u}\implies\frac{h_i}{2}=-\frac{-30}{-15}\)
• \(h_i=-4cm\)

So, value of X = 30 cm and h = 4 cm.

Hence, the correct answer is X = 30 cm; h = 4 cm.

Concept:

Myopia:

• Myopia or Nearsightedness occurs when the eye loses its ability to focus on far-off objects as the lenses do not possess a long focal length.
• Objects that are near are clearly visible to patients with this defect.
• As we know, from the reference for countless ray diagrams describing the functioning of the eye, when the light suffers higher refraction than usual, the eye would not be able to form an image for faraway objects.

​

• This eye defect is corrected with a concave lens.

​Explanation:

• Myopia is corrected by using concave lenses of suitable focal length (or power).
• So that it can produce an additional diversion in the light rays and the final image is formed on the retina.

Additional InformationHypermetropia:

• Hypermetropia is also referred to as hyperopia or long-sightedness, or far-sightedness.
• Hypermetropia is the condition of the eyes where the image of a nearby object is formed behind the retina.
• Here, the light is focused behind the retina instead of focusing on the retina.

• Hypermetropia is mainly caused due to certain structural defects in the retina. Structural defects include:

  • Small-sized eye-ball
  • Non-circular lenses
  • The cornea is flatter than usual
• A convex lens is used for the correction of hypermetropia.

The correct answer is a - (iv), b - (i), c - (v), d - (ii), e - (iii).

Concept:

• The human eye is one of the most valuable and sensitive sense organs.
• It enables us to see the wonderful world and the colors around us. On closing the eyes, we can identify objects to some extent by their smell, taste, sound they make, or by touch.
• It is, however, impossible to identify colors while closing the eyes. Thus, of all the sense organs, the human eye is the most significant one as it enables us to see the beautiful, colourful world around us.

Explanation:

Parts of an eye:
The adult human eyeball is nearly spherical in structure. It consists of tissues present in three concentric layers
i) Outermost fibrous layer composed of sclera and cornea.
ii) Middle layer consists of choroid, ciliary muscles and iris.
iii) Innermost layer consists of retina

Outermost layer-

• Sclera:
  • It is an opaque outermost layer, composed of dense connective tissue that maintains the shape of the eyeball and protects all the inner layers of the eyeball.
• Cornea:
  • The cornea is a thin transparent, front part of sclera, which lacks blood vessels, but is rich in nerve endings. These nerve endings are in charge of the involuntary reflex of closing the eyelid in response to touch, preventing foreign particles from entering and causing damage to the eye.
  • The cornea and lens work together to help in focusing light rays on the back of the eye.

Middle layer-

Choroid:

• It is a pigmented layer (bluish) present beneath the sclera. It contains numerous blood vessels and nourishes the sclera.
• The choroid layer is thin over the posterior two-thirds of the eyeball, but it becomes, thick in the anterior part to form the ciliary muscles.

Ciliary muscles:

• Ciliary muscles holds the lens in position, stretching and relaxation of ciliary muscles changes the focal length of the lens for accommodation.

Iris:

• Iris forms a pigmented circle of muscular diaphragm attached to the ciliary body in front of the lens.
• The pigment present in it gives characterstic colour to the eyes.
• Iris controls the amount of light that enters the eye by adjusting its size. The iris is a thin membrane that controls the pupil, which in turn controls the amount of light that enters the eye.

Pupil:

• It is the aperture surrounded by the iris.
• The cornea bends light rays so that they pass freely through the pupil, which allows light into the eye.

Innermost layer

• Retina:
  • The retina is a delicate membrane having enormous number of light-sensitive cells. 
  • The light-sensitive cells get activated upon illumination and generate electrical signals.
  • Retina acts as screen for the image.

Concept:

• The Refractive index is the ratio between the speed of light in air to the speed in a medium.
• Snell’s law clarifies the relation between the angle of incidence(i) and angle of refraction(r).
• \(μ = \frac {sin ~r }{sin~ i}\)
• Area of circle = πr²

Calculation:

Given that, μ = \(\frac{2}{\sqrt3}\) , depth of tank = 80 cm = 0.8 m

When light is placed at the centre of the bottom then the light come out in conical shape

Now let the angle OBC = i and after refraction angle (r) = 90° 

Then from Snell's law \(μ = \frac {sin ~r }{sin~ i}\)

\(\frac 2 {√ 3} = \frac {sin~ 90}{sin~ i}\)

⇒ i = 60° 

Now, in triangle OBC

tan 60° = \(\frac {OC}{OB} = \frac {OC}{0.8}\)

⇒ OC = √3 × 0.8

The area of the surface of the liquid through which light from the source can emerge out is circular and radius is equal to OC

Then the area of the surface = πr² = π(√3 × 0.8)² = 6.03 m²

The correct answer is 0.75 m.

Key Points

• To see one's full image in a vertical plane mirror, a person's eye level must be at the halfway point between the top and bottom of the mirror.
  • This means that the minimum height of the mirror should be half of the height of the person.
  • This way, the complete image of the person will be visible in the mirror.
• Height of Sita is 1.5 meters
• The mirror's height should be 1.5/2 = 0.75 m