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

KEY: b = 8 × 105

Solution:

The focal length f3 of the liquid lens in air is calculated by:

1/f3 = (μ - 1) (1/R2 - 1/R3)

Given:

μ = 1.4, R2 = -25 cm, R3 = +60 cm

1/f3 = (1.4 - 1) (1/(-25) + 1/60)

⇒ 1/f3 = 0.4 (-1/25 + 1/60)

⇒ 1/f3 = -7 / 750

So, f3 ≈ -107.14 cm

The formula for the equivalent focal length is:

1/F = 1/f1 + 1/f2 + 1/f3

Given: f1 = 25 cm, f2 = 60 cm, f3 ≈ -107.14 cm

1/F = 1/25 + 1/60 - 1/107.14

⇒ 1/F ≈ (60 + 25 - 14) / 1500

⇒1/F ≈ 71 / 1500

So, F ≈ 1500 / 71 ≈ 21.13 cm

Using the lens formula

1/v - 1/u = 1/F

Object distance u = -80 cm, so:

1/v - 1/(-80) = 1/21.13

⇒1/v = 1/21.13 - 1/80

v ≈ 28.7 cm

Calculation:

The formula for the total power of lenses in contact is:

P = Pconvex + Pconcave

Pconvex = 3 D

focal length of concave lens = -50 cm = -0.5 m

Power of concave lens:

Pconcave = 1 / (-0.5) = -2 D

Total power:

P = 3 + (-2)

⇒ P = 1 D

The formula for the equivalent focal length is:

P (in D) = 1 / f (in m)

So, f = 1 / P

⇒ f = 1 / 1 = 1 m

the ratio = P/f =1 D²

Calculation:

Using the lens formula:

1/v1 - 1/u1 = 1/f1

Where, u1 = -24 cm (object distance from the lens) and f1 = 12 cm,

1/v1 - 1/(-24) = 1/12

⇒ 1/v1 + 1/24 = 1/12

⇒ v1 = 24 cm

I1 is formed 24 cm from the lens, so the distance from the mirror is 24 - 8 = 16 cm to the right of the mirror.

This image acts as a virtual object for the mirror, so the mirror forms an image at the same distance on the other side:

So, the mirror image is located 16 cm left of the mirror.

Now, this mirror image acts as an object for the lens. Distance from the lens:

Lens to mirror = 8 cm, mirror to image = 16 cm, so distance from lens = - (16 + 8) = -24 cm (since it is to the left of the lens).

Apply the lens formula again:

1/v2 - 1/(-24) = 1/(-12) (negative because light is incident from the right)

⇒ 1/v2 + 1/24 = -1/12

⇒ v2 = -8 cm

The negative sign shows that the image is on the left side of the lens, meaning it is real. The distance of this final image from the mirror:

Distance between lens and mirror = 8 cm. Image is 8 cm to the left of the lens, so total distance from mirror = 8 + 8 = 16 cm.

Thus, the final image is real and located 16 cm from the mirror.

Explanation:

(A) Light intensity depends on the aperture area, hence radius of the objective lens.

(B) Angular magnification of a telescope is given by the ratio of focal lengths of objective to eyepiece.

(C) Telescope length is approximately the sum of focal lengths of objective and eyepiece.

(D) Image clarity is improved by minimizing spherical aberrations.

The correct answer is Scattering.

Key Points

• When light passes from one medium to any other medium say air, a glass of water then a part of the light is absorbed by particles of the medium preceded by its subsequent radiation in a particular direction. This phenomenon is termed a scattering of light.
• Rayleigh's law of scattering:-
  • According to Rayleigh's law of scattering, the intensity of light of wavelength λ present in the scattered light is inversely proportional to the fourth power of λ, provided the size of the scattering particles is much smaller than λ. Mathematically,
  • \(I \propto \frac{1}{\lambda^4 }\)
• Thus the scattered intensity is maximum for a shorter wavelength.

​Additional Information

• The molecules of air and other fine particles in the atmosphere have a size smaller than the wavelength of visible light.
• These are more effective in scattering light of shorter wavelengths at the blue end than the light of longer wavelengths at the red end.
• The red light has a wavelength of about 1.8 times greater than blue light.
• Thus, when sunlight passes through the atmosphere, the fine particles in the air scatter the blue colour (shorter wavelengths) more strongly than red. 
• The scattered blue light enters our eyes. 
• The red colour of the sun at sunrise is due to the scattering of light. 

CONCEPT:

• Power of the Lens: The lens power is the inverse of the focal length in meters.
  • The physical unit for lens power is 1/meter which is called dioptre (D).

Power (P) = 1/f 

Where f is the focal length of the lens

CALCULATION:

Given that:

Focal length of the lens (f) = 1 cm = 1/100 = 0.01 m

So the power of the lens, \(p\; = \;\frac{1}{{f}}\; = \;\frac{1}{0.01}\; = \;100\;D\)

CONCEPT:

• Lens formula: The expression which shows the relation between object distance (u), image distance (v), and focal length (f) is called lens formula.

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

Linear magnification (m):

• It is defined as the ratio of the height of the image (h_i) to the height of the object (h_o).

i.e. \(m = \frac{{{h_i}}}{{{h_o}}}\)

• The ratio of image distance to the object distance is called linear magnification.

i.e. \(m = \frac{{image\;distance\;\left( v \right)}}{{object\;distance\;\left( u \right)}} = \frac{v}{u}\)

CALCULATION:

Given: u = - 30 cm and f = 15 cm

Lens formula

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

\(\therefore \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{{15}} - \frac{1}{{30}} = \frac{{2 - 1}}{{30}} = \frac{1}{{30}}\;cm\)

v = 30 cm

Linear magnification (m)

\(m = \frac{v}{u}\)

\(\Rightarrow m = \frac{{30}}{{ - 30}} = - 1\)

The correct answer is Real and inverted.

• The image formed by the concave mirror when the object is placed beyond the center of curvature is Real and Inverted.

Key Points

• Concave Mirror:
  • A spherical mirror, whose reflecting surface is curved inwards, that is, faces towards the center of the sphere, is called a concave mirror.
  • Concave mirrors are commonly used in  Torches, Search-lights, Vehicle headlights, Shaving mirrors, Microscopes, and Telescopes.
  • Image formation by a concave mirror for different positions of the object:

The correct answer is Concave mirrors.

Key Points 

• Concave mirrors are spherical mirror in which the reflecting surface curved inwards.
• A concave mirror forms a virtual or real image depending on the position of the object.
• Concave mirrors are used by dentists to see an enlarged image of the teeth.
• Other important uses of Concave mirrors are:
  • ​Used in torches.
  • Used in solar cooker.
  • Used in search-lights and vehicles headlights to get powerful parallel beams of light.
  • Used as shaving mirrors to see a larger image of the face.
• Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Additional Information 

• Convex mirrors are the spherical mirrors in which the reflecting surface curved outwards.
• The important uses of a convex mirror are:
  • Used as rear-view mirrors in vehicles.
  • Used for Security purposes.
• Plane mirrors are used in Periscopes and kaleidoscopes.

CONCEPT:

• Plane Mirror: A plane mirror is a mirror with a flat (planar) reflective surface.

The characteristics of an image formed in a plane mirror:

• The image formed by the plane mirror is virtual and erect i.e. image cannot be projected or focused on a screen.
• The distance of the image ‘behind’ the mirror is the same as the distance of the object in front of the mirror.
• The size of the image formed is the same as the size of the object.
• The image is laterally inverted, i.e. left hand appears to be right hand when seen from the plane mirror.

• If the object moves towards (or away from) the mirror at a certain rate, the image also moves towards (or away from) the mirror at the same rate.

EXPLANATION:

From the above discussion, we can say that,

• The image formed by Plane Mirror is always Virtual and Erect. So option 1 is correct.
• A convex lens and a concave mirror forms both real and virtual images.
• A concave lens and a convex mirror can form only virtual images.

The correct answer is at the centre of the curvature.

Key Points

Position of the Object at the Centre of Curvature: 
When an object is placed at the center of curvature (C) of a concave mirror, the object is essentially at a distance equal to the radius of curvature from the mirror's pole.
Formation of Image:

• In this scenario, the reflected rays from the object converge and meet at the center of curvature.
• The image formed is real because the reflected rays actually converge at a point.
• The image is inverted because the reflected rays cross over at the center of curvature, resulting in an inverted orientation.
• The image is of the same size as the object because the distance from the object to the mirror is the same as the distance from the image to the mirror.

• If an object is placed at the centre of curvature of a concave mirror, then the image formed is real, inverted, and of the same size as that of the object.

CONCEPT:

Concave mirror: The mirror in which the rays converge after falling on it is known as the concave mirror.

• Concave mirrors are also known as a converging mirror.
• The focal length of a concave mirror is negative according to the sign convention.

Magnification: In a concave mirror, the magnification is the ratio of the height of the image to the height of the object.

When the image is real, the magnification will be negative because the real image is inverted.

• When the image is virtual, the magnification will be positive because the virtual image is erect.

 \(m = \frac{-v}{u}\)

where m is magnificent, v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

• Mirror Formula: The following formula is known as the mirror formula:

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

where f is focal length v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

CALCULATION:

Given that u = -20 cm and m = -3.

\(m = \frac{-v}{u} \)

\(-3=\frac{-v}{-20}\)

v = -60 cm

Mirror formula \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

\(\frac{1}{f}=\frac{1}{-20}+\frac{1}{-60}\)

f = -60 / 4= -15 cm

So the correct answer is option 1.

Compound microscope: 

• It is used for much larger magnifications.
• Two lenses are used with one lens compounding the effect of the other.

where fo is the focal length of the objective lens,  fe  is the focal length of the eyepiece, h is the object height, h' is the size of the first image

• A compound microscope uses two lenses i.e., objective lens and eyepiece. 
  • Objective lens: lens nearest to the object forming a real, inverted, and magnified image of the object.
  • Eyepiece: produces a final image that is enlarged and virtual.

​CALCULATION:

Given: magnification of compound microscope(m) = 20, focal length of eye piece (f_e) = 5 cm and image is formed at near point (v= D) = 25 cm.

• Linear magnification due to eyepiece:

\(⇒ m_{e} = 1+\frac{D}{f} = 1+ (\frac {25}{5} )= 6\)

• Total magnification is given by:

⇒ m = m_o × m_e,

where m_o is the magnification due to the objective lens.

• Magnification due to the objective lens is given by:

\(⇒ m_{o} = \frac {m}{m_{e}} = \frac {20}{6} = 3.33\)

• Hence option 3) is correct.

Concept:

Refraction of Light: When a ray of light is traveling from one transparent medium to another, it bends its path. This phenomenon is called refraction. 

When the light changes its medium, its speed, and wavelength changes.

​

Where i is the angle of incidence, r is the angle of refraction.

Refractive Index: 

The ratio of the speed of light in the vacuum to the speed of light in a given transparent medium is called the refractive index of the medium.

• When light travels from a medium with a higher refractive index to a lower one, then it bends away from normal. The angle of refraction is more than the angle of incidence.
• When light travels from a medium with a lower refractive index to a higher one, then it bends toward normal. The angle of refraction is less than the angle of incidence.

Snells Law of Refraction: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for two transparent medium.

\(\frac{sin\; i}{sin\; r} = \;\frac{n_2}{n_1}\)

n₂ is the medium in which light is entering, n₁ is the initial medium of light

When the ray of light is traveling from the medium with a higher refractive index to a lower one, for example, water to air, and the angle of refraction formed is 90 ° then the angle of incidence at that point is critical angle.

​

When medium 2 is air having refractive index n₂ = 1, and medium 1 have refractive index n₁ =  μ, then critical angle θc can be represented by 

\(\frac{sin\; θ_c}{sin \;90 °} = \frac{1}{μ}\)

\(\implies sin\; θ_c= \frac{1}{μ}\)

• The sine of critical angle is reciprocal of refractive index.
• When the incidence angle is more than the critical angle, then the ray of light reflects back in the same medium. This phenomenon is called Total Internal Reflection.

Calculation:

Given refractive index μ = 1. 62

\(sin\; θ_c= \frac{1}{μ}\)

\(sin\; θ _c= \frac{1}{1.62}\)

sin θ_c = 0. 62

So, 0.62 is the correct option.