Refraction and Reflection MCQ Quiz - Objective Question with Answer for Refraction and Reflection - Download Free PDF

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.

Calculatiion:

δ_net = 0

(μ₁ – 1)A₁ – (μ2 – 1)A₂ = 0

(1.54 – 1)4 – (1.72 – 1)A₂ = 0

A₂ = 3°

Calculation:

\(\sin \mathrm{c}=\frac{1}{μ}\)

for μ → least, c → maximum

θ = c = 45

\(\mu=\frac{1}{\sin 45}=\sqrt{2}\)

Calculation: 

Maxima condition

\(2 μ \mathrm{t}=\mathrm{n} λ \Rightarrow \mathrm{t}=\frac{\mathrm{n} λ}{2 μ} \Rightarrow \mathrm{t}=\frac{λ}{2 μ}, \frac{2 λ}{2 μ}, \ldots \ldots\)

Minima condition 2μt = (2n – 1)λ/2

⇒ \(\mathrm{t}=\frac{(2 \mathrm{n}-1) \lambda}{4 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{4 \mu}, \frac{3 \lambda}{4 \mu}, \ldots .\)

\(\Delta t=\frac{2 \lambda}{4 \mu}\)

Rate of evaporation = \(\frac{\mathrm{A}(\Delta \mathrm{t})}{\text { time }}=54 \times 10^{-13} \mathrm{~m}^{3} / \mathrm{s}\)

​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:

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 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.

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:

• When light travels from air into glass slab, it slows down, causing it to change direction slightly. This change of direction is called refraction.
• When light enters a denser substance (higher refractive index), it ‘bends’ more towards the normal line.

• When a ray of light enters a denser medium from rare medium then its speed decreases as it encounters a dense medium.
• Since to conserve energy the frequency must remain constant, the wavelength of light changes.

The relation between speed of light, wavelength and frequency

• c = μ λ 
• Where, c is the speed of light, μ = frequency of light, λ = wavelength.

Explanation:

Since the glass slab is the denser medium than the air, then the speed of light decreases and bends towards the normal.

As we know the that the frequency of light is not change, 

Then, c ∝ λ 

If the speed of the light decreases, then the wavelength of light also decreases.

Wavelength is decreased by a factor called refractive index of the medium.

Calculation:

The formula for the refractive index is given as:

r₁ + c = A

⇒ r₁ = 90 - c

Snell's law at the emergent boundary

Sin(c) = 1 / μ

⇒ Cos(c) = (μ² - 1)^1/2 / μ

According to Snell's law at incident boundary

Sin(30) = μ × Sin(r₁)

1/2 = μ × Sin(90 - c) =μ Cos(c)

⇒ 1/2 = μ × (μ² - 1)^1/2 / μ

Squaring both sides:

1/4 = μ² - 1

μ² = 5 / 4

μ = √(5 / 4) = √5 / 2

Thus, the correct option is option 2: μ = √5 / 2

Concept:

• For a plano-convex lens, one of the surfaces is flat (Plano) and the other is curved (convex).
• The radius of curvature of the curved surface is given as 20 cm, and the refractive index of the material of the lens is 1.5.
• The focal length of the lens can be calculated using the lens maker's formula:

1/f = (n - 1) × (1/R1 - 1/R2)

where

f is the focal length of the lens,

n is the refractive index of the lens material,

R1 is the radius of curvature of the first surface (the flat surface in this case, which is infinite), and

R2 is the radius of curvature of the second surface (the curved surface in this case).

Since the first surface is flat, the radius of curvature R1 is infinite, and the term (1/R1) becomes zero. Therefore, the lens maker's formula simplifies to:

1/f = (n - 1) ×  (1/R2)------(1)

Calculation:

Substituting the values in equation (1), we get:

1/f = (1.5 - 1) × (1/20)

1/f = 0.025

f = 1/0.025

f = 40 cm

Therefore, the focal length of the plano-convex lens is 40 cm.

The correct answer is option (4)

Concept:

• lens is a transmissive optical device that focuses or disperse light beam using refraction.
• Two types of lenses: Concave lens and convex lens

Concave lens: 

• It is flat in the middle and thicker at the edge.
• It is also known as a diverging lens because it bends the parallel light rays outwards and diverges them at the focal point. 

Convex lens: 

• It is thick in the middle and thinner on the edge.
• This lens is also known as a converging lens because it converges all light at a point called the focus.

. 

• The working principle of the lens is the law of refraction.
• The total power of the two lenses when placed in contact, P = P1 + P2
• The power of the lens is inversely proportional to its focal length, \(P=\frac {1}{f}\)

Lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) where v = image distance, u = image distance, f = focal length

Explanation:

Explanation:

Given, Power, P₁(L₁) = 10D , P₂ (L₂) = 12.5 D

The distance between the two lenses, d = 44 cm

For the first convex lens L₁, f = 1/P =100/10 = 10 cm

Given, u = - 20 cm, f = 10 cm

Lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

\(\frac{1}{10}=\frac{1}{v}-\frac{1}{-20}\)

v = +20 cm

For the second convex lens L₂, f = 1/P =100/12.5 = 8 cm

Object distance, u = - 44 - (-20) = - 24 cm, focal length, f = 8 cm

Lens formula, \(\frac{1}{8}=\frac{1}{v}-\frac{1}{-24}\)

v = 12 cm

The distance between the original object and the final image is 

= 20 + 44 + 12 = 76 cm

CONCEPT:

• Lens: The transparent curved surface which is used to refract the light and make an image of any object placed in front of it is called a lens.
  • Convex lens: ​A lens having two spherical surfaces, bulging outwards is called a double convex lens (or simply convex lens).
  • It is thicker in the middle as compared to the edges.
  • Convex lenses converge light rays and hence, convex lenses are also called converging lenses.

• Lens formula is given by:

\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

• The magnification of the lens is given by:

Magnification (m) = -v/u

Where u is object distance, v is image distance and f is the focal length of the lens

• A negative magnification implies an inverted image.
• In the case of a convex lens, the magnification is -1 when the object is placed at 2F (twice the focal length).
• This is because at this position, the image is formed on the opposite side of the lens, at the same distance as the object, but inverted.