Thin Lenses MCQ Quiz - Objective Question with Answer for Thin Lenses - Download Free PDF

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.

Calculation:

Using lens maker's formula for original lens:

(μ − 1) × (1/R₁ − 1/R₂) = (μ − 1) × (2/R)

If the lens is replaced by another lens with unequal radii R₁ and R₂, then to maintain the same lens power:

1/R₁ + 1/R₂ = 2/R = 2 × (1/6) = 1/3

Now, check which pair satisfies 1/R₁ + 1/R₂ = 1/3:

Option (2): R₁ = 1/5 cm, R₂ = 1/7 cm

1/5 + 1/7 = (7 + 5)/35 = 12/35 ≈ 0.343

1/3 ≈ 0.333 ⇒ Close match

Thus, best approximate valid answer among options is:

Final Answer: R₁ = 1/5 cm and R₂ = 1/7 cm

Hence, the correct option is (2).

Explanation:

The lens formula is: 1/v - 1/u = 1/f

Given that the focal length (f) is 10 cm, and the image distance (v) is 15 cm (positive for a real image):

1/15 - 1/(-u) = 1/10

⇒ 1/u = 1/10 - 1/15

⇒ 1/u = (3 - 2) / 30

⇒ 1/u = 1/30

Thus, u = 30 cm

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, \(p\; = \;\frac{1}{{f}}\; = \;\frac{1}{0.01}\; = \;100\;D\)

Option 2 is the correct answer: 

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.
• Concave lens: It is a diverging lens that diverges the parallel beam of light.
  • It can also gather light from all directions and project it as a parallel beam.
  • It has a virtual focus from the diverging rays of light that seem to converge.
  • The image formed by a concave lens is virtual, erect, and diminished.

The lens equation which gives the relation between focal length, the distance of image and object from the lens is

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

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

CALCULATION:

• The focal length of a concave lens is taken negative (is measured opposite to the incident rays).

Given that:

focal length f = -15 cm

Distance of object u = - 30 cm

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

substituting these values in the lens equation we get,

\({1 \over v} = {-{1 \over 15}-{1 \over 30}}\)

\({1 \over v}={-2-1 \over 30}\)

\({1 \over v}={-3 \over 30}\)

\(v=-10cm\)

Hence distance of the object from the lens is 10 cm and the negative sign signifies a virtual image. So option 2 is correct.

EXTRA POINTS:

Positions of the object and image for the concave lens:

Concept:

• The focal length of a combination of lenses is calculated by:

\(\frac{1}{f} = \frac{1}{f_1} +\frac{1}{f_2}\)

Calculation:

Given f₁ = + 50 cm (For convex lens) and f₂ = - 40cm (For concave lens)

\(\frac{1}{f} = \frac{1}{50} + \frac{1}{-40}\)

\( ⇒\frac{1}{f} = \frac{4-5}{200}\)

\( ⇒ \frac{1}{f} = -\frac{1}{200}\)

⇒ \(f = - 200\ cm\)

• Since focal length is negative therefore the combination of the lens is a diverging lens of 200 cm focal length.

 Additional Information

Power is reciprocal to focal length.

\(P = \frac{1}{f}\)

The correct answer is the Concave lens.

Key Points

• Myopia
  • It is an eye disorder of near-sightedness.
  • In myopia, light focuses in front of, instead of on, the retina.
  • This makes distant objects blurry and close objects appear normal.
• Concave lens
  • A concave lens is a lens that possesses at least one surface that curves inwards.
  • It is a diverging lens
  • It is thinner at its centre than at its edges.
  • Uses of Concave lens - in telescopic, in Eye Glasses, etc.

Additional Information

• Convex lens
  • It is a lens that converges rays of light that are travelling parallel to its principal axis.
  • It is a converging lens.
  • It is thicker at the centre and thinner at the edges.
  • It is used to correct Hypermetropia, used in cameras.
• Cylindrical lens
  • A cylindrical lens is a lens that focuses light into a line instead of a point
  • The lens compresses the image in a direction perpendicular to this line.
  • Cylindrical lenses are used in optical spectrometers.
  • It is used in holography.

CONCEPT:

• 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.
• The lens whose refracting surface is upside is called a convex lens and the lens whose refracting surface is inside is called a concave lens or diverging lens.
  • Concave lenses diverge light rays and hence are also called diverging lenses.

Positions of the object and image:

EXPLANATION:

• When the object is between the infinity and optical center of the lens then the image forms between focus and optical center which is diminished, virtual and erect. So option 1 is correct.

CONCEPT:

• Power of Lens: The inverse of the focal length is known as the power of the lens.
  • It shows the bending strength for the light ray of the lens.
  • The unit of power of a lens is Dioptre when the focal length of the lens is taken in meter (m).

\(P = \frac{1}{f}\)

where P is the power of the lens and f is the focal length of the lens.

• Concave lens: It is a diverging lens that diverges the parallel beam of light.
  • It can also gather light from all directions and project it as a parallel beam.
  • The focal length of the concave lens is negative.
  • It has a virtual focus from the diverging rays of light that seem to converge.
• Convex lens: The lens whose refracting surface is upside is called a convex lens.
  • The convex lens is also called a converging lens.
  • The focal length of a convex lens is positive.

CALCULATION:

Focal length (f) = 50 cm = 50/100 = 0.5 m

Since f = 1/P 

So power of the lens (P) = 1/f = 1/0.5 = 2 D

The focal length of a convex lens is positive. 

So the correct answer is option 3.

CONCEPT:

• Power of Lens: The inverse of the focal length is known as the power of the lens.
  • It shows the bending strength for the light ray of the lens.
  • The unit of power of a lens is Dioptre when the focal length of the lens is taken in meter (m).

\(P = \frac{1}{f}\)

where P is the power of the lens and f is the focal length of the lens.

• Concave lens: It is a diverging lens that diverges the parallel beam of light.
  • It can also gather light from all directions and project it as a parallel beam.
  • The focal length of the concave lens is negative.
  • It has a virtual focus from the diverging rays of light that seem to converge.
• Convex lens: The lens whose refracting surface is upside is called a convex lens.
  • The convex lens is also called a converging lens.
  • The focal length of a convex lens is positive.

CALCULATION:

Given that P = 1.5 D 

Focal length (f) = 1/P = 1/(1.5) = 1/1.5 = 0.666 m = 66.6 cm

The focal length of a convex lens is positive.

So the correct answer is option 2.

CONCEPT:

• 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.
  • The lens whose refracting surface is upside is called a convex lens.
  • The convex lens is also called a converging lens.

Lens formula is given by:

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

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

CALCULATION:

Given that: 

Object distance (u) = - 20 cm

Focal length (f) = 7.5 cm

Use the lens formula:

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

\(\frac{1}{v} - \frac{1}{-20} = \frac{1}{7.5}\)

\(\frac{1}{v} = \frac{1}{7.5} - \frac{1}{20} =\frac{20-7.5}{7.5\times 20}=\frac{12.5}{150}\)

So image distance (v) = 150/12.5 = 12 cm

Hence option 3 is correct.

Concept:

There are mainly 2 types of lenses:

Concave lens:

• It is a type of lens that possess at least one surface that is curved inside.
• It is also known as the diverging lens.

The characteristics of an image formed in a Concave Lens:

• The concave lens is thinner in the middle and thicker at the edge.
• The image formed by a concave lens is always upright, virtual, and erect.
• The size of the image is always smaller than the object.
• The image can be found between the focus and the lens.
• The focal length is always negative.

Convex Lens:

• It is a type of lens that possess at least one surface which curves outside much similar to the exterior of a sphere.
• It is also known as a converging mirror.

The characteristics of an image formed in a Convex Lens:

• ​A convex lens is thicker in the middle and thinner at the edge.
• Depending on the point of the object:     
  • The image may be real or virtual.
  • image may be erect or inverted.
  • image may be magnified or diminished.

Explanation:

• A concave lens diverges incoming light rays and always produces an upright, diminished and virtual, and erect image of the object in front of it.

Image formation by the concave lens:

Thus, from the above discussion, we can say that the correct answer is It is upright and smaller than the object.

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: The lens whose refracting surface is upside is called a convex lens.
  • The convex lens is also called a converging lens.

Lens formula is given by:

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

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

CALCULATION:

Given that: 

Object distance (u) = - 20 cm

Focal length (f) = 7.5 cm

Use 

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

\(\frac{1}{v} - \frac{1}{-20} = \frac{1}{7.5}\)

\(\frac{1}{v} =- \frac{1}{20} + \frac{1}{7.5} =1/12\)

So v = 12 cm.

Hence option 3 is correct.