Convex Lens Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. What is a lens?

Ans. A lens is an optical device that converges or diverges the beam of light by means of refraction. It consists of a piece of transparent material bounded by two refracting surfaces out of which at least one is curved.

Ques. A simple telescope has

1. A concave mirror and a convex lens
2. A convex lens and a concave lens
3. Two concave lenses
4. Two convex lenses

Ans. The correct answer is d. Two convex lenses

Explanation: Telescope is an optical instrument used to clearly observe a distant object. A simple telescope consists of two convex lenses mounted co-axially in two metallic tubes.

Ques. Which of the following is a converging lens?

1. A mirror
2. Concave lens
3. Convex lens
4. Planar lens

Ans. The correct answer is c. Convex lens

Explanation: A Convex lens converges the incident parallel beam of light to its principal focus, that's why it is also known as a converging lens.

Ques. A prime lens has _____.

1. fixed focal lengths
2. varying focal lengths
3. neither fixed nor varying focal lengths
4. both fixed and varying focal lengths

Ans. The correct answer is a. fixed focal length

Explanation: A prime lens is a fixed focal length photographic lens, typically with a maximum aperture from f2.8 to f1.2.

Ques. Which type of lens is used in a magnifying glass?

1. Planar lens
2. Concave lens
3. Convex Lens
4. A mirror

Ans. The correct answer is c. Convex Lens

Explanation: Magnifying glass is an optical device used to see tiny objects clearly. Convex lenses are used in magnifying glasses.

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Short Answers Questions [2 Marks Questions]

Ques. What is the difference between a concave and convex lens?

Ans. The following are the differences between a concave and convex lens

Ques. Define the focal length of a lens.

Ans. The distance between the optical center and the principal focus of a lens is called the focal length. It is denoted by f.

Focal length of a convex lens is taken to be positive and the focal length of a concave lens is taken to be negative.

Ques. What are the uses of convex lenses?

Ans. The following are the uses of a convex lens

• Convex lenses are used in eyeglasses of a person who is suffering from farsightedness or hypermetropia. 
• It is used in microscopes to magnify tiny objects or microorganisms such as bacteria, microbes, etc.
• Convex lenses are used in cameras to focus and magnify images.
• These are used in telescopes to see far objects clearly.

Ques. Write the lens formula for a convex lens.

Ans. The lens formula is given by

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

Where

• f is the focal length of the lens
• u is the distance of the object from the lens
• v is the distance of the image from the lens

This formula is valid for both convex and concave lens.

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Long Answers Questions [3 Marks Questions]

Ques. The radius of curvature of the faces of a double convex lens is 10 cm and 15 cm. If the focal length of the lens is 12 cm, find the refractive index of the material of the lens.

Ans. Given

• Radius of curvature of the first refracting surface of the lens, R₁ = 10 cm
• Radius of curvature of the second refracting surface of the lens, R₂ = - 15 cm
• Focal length of the lens, f = 12 cm

Using the lens maker formula,

\(\frac{1}{f} = [\mu - 1] [\frac{1}{R_1} - \frac{1}{R_2}]\)

⇒ \(\frac{1}{12} = [\mu - 1] [\frac{1}{10} + \frac{1}{15}] = \frac{\mu - 1}{6}\)

⇒ \(\frac{1}{12} = \frac{\mu - 1}{6}\)

⇒ μ = \(\frac{3}{2}\) = 1.5

Ques. A thin converging lens made of glass with a refractive index of 1.5 acts as a concave lens of focal length 50 cm when immersed in a liquid of refractive index 15/8. Calculate the focal length of the converging lens in air.

Ans. When the converging lens is immersed in liquid it acts as a concave lens. Therefore its focal length is taken to be -ve.

Given

• Focal length of the lens in liquid, f_l = - 50 cm
• Refractive index of the lens with respect to air, µ^a_g = 1.5
• Refractive index of liquid with respect to air, µ^a_l = 15/8 = 1.86

Let f_a be the focal length of the lens in air, then using the formula

\(\frac{f_l}{f_a} = \frac{\mu_g^a-1}{(\frac{\mu_g^a}{\mu_l^a}-1)}\)

⇒ \(\frac{-50}{f_a} = \frac{1.5-1}{(\frac{1.5}{1.86})-1}\)

⇒ f_a = 19.36 cm

Hence the focal length of the converging lens in air is 19.36 cm.

Ques. The focal length of an equi-convex lens is equal to the radius of curvature of either face. What is the refractive index of the material of the lens?

Ans. Let R be the radius of curvature of the faces of the lens, then

• Focal length of the lens, f = R
• Radius of curvature of the first refracting face of the lens, R₁ = +R
• Radius of curvature of the second refracting face of the lens, R₂ = -R

Using the lens maker formula

\(\frac{1}{f} = [\mu - 1] [\frac{1}{R_1} - \frac{1}{R_2}]\)

⇒ \(\frac{1}{R} = [\mu - 1] [\frac{1}{R} + \frac{1}{R}] = \frac{2(\mu - 1)}{R}\)

⇒ 2μ – 2 = 1

⇒ μ = \(\frac{3}{2}\) = 1.5

Hence the refractive index of the material of the lens is 1.5

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Very Long Answers Questions [5 Marks Questions]

Ques. Derive lens maker formula using a convex lens.

Ans. Consider a convex lens of focal length f, and R₁ and R₂ be its radii of curvature.

For the first refracting surface

\(- \frac{\mu_1}{u} + \frac{\mu_2}{v_1} = \frac{\mu_2 - \mu_1}{R_1}\)……(i)

For the second refracting surface

\(- \frac{\mu_2}{v_1} + \frac{\mu_1}{v} = \frac{\mu_1 - \mu_2}{R_2}\) ……(ii)

Adding both the equations (i) and (ii), we get

 \(- \frac{\mu_1}{u} + \frac{\mu_1}{v} = (\mu_2 - \mu_1) [\frac{1}{R_1}- \frac{1}{R_2}]\)

⇒ \(- \frac{1}{u} + \frac{1}{v} = [\frac{\mu_2}{\mu_1}-1] [\frac{1}{R_1}- \frac{1}{R_2}]\)

But µ₂ / µ₁ = µ (relative refractive index of the lens with respect to air or rarer medium).

⇒ \(- \frac{1}{u} + \frac{1}{v} = [\mu-1] [\frac{1}{R_1}- \frac{1}{R_2}]\)

if u = - ∞ then v = f

⇒ \(- \frac{1}{-\infty} + \frac{1}{f} = [\mu-1] [\frac{1}{R_1}- \frac{1}{R_2}]\)

Since 1/∞ = 0, therefore we can write

⇒ \(\frac{1}{f} = [\mu-1] [\frac{1}{R_1}- \frac{1}{R_2}]\)

The above equation is known as lens maker formula.

Ques. The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distance.

Ans. Let the size of the object = h

Then the size of the image, h’ = 4h

Therefore, the magnification of the lens, m = h’/h = 4h/4 = 4

Also the magnification of a lens is given by, m = v/u

Where

• v is the image distance
• u is the object distance

⇒ v/u = 4 

⇒ v = 4u

Now using the lens formula, 1/f = - 1/u + 1/v

⇒ 1/20 = - 1/u + 1/4u = - 3/4u

⇒ 4u = - 60

⇒ u = -15 cm

Substituting the value of u in equation v = 4u, we get

v = 4 x (-15) = - 60 cm

Ques. A convex lens of focal length 30 cm and a concave lens of focal length 60 cm are placed in combination. If the object is placed 40 cm away from the combination, find the position of the image.

Ans. Given

• The focal length of the convex lens, f₁ = + 30 cm
• The focal length of the concave lens, f₂ = – 60 cm

The focal length of the equivalent lens is given by

1/f = 1/f₁ + 1/f₂ = 1/30 - 1/60

⇒ f = + 60 cm

So the focal length of the equivalent lens is + 60 cm. It acts as a convex lens.

Now using, 1/f = – 1/u + 1/v

⇒ 1/60 = - 1/40 + 1/v

⇒ v = - 120 cm

Hence, the position of the image is 120 cm in front of the lens.

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