Engineering Physics Questions and Answers – Lenses – 1

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Lenses – 1”.

1. What is the lens?
a) An image – forming device
b) An image – producing device
c) An image – reflecting device
d) An object – reflecting device
View Answer

Answer: a
Explanation: A lens is an image- forming device. They form an image by refraction of light at its two bounding surfaces.

2. The spherical surface of lens results in ________
a) having a wide range of focal length
b) having a narrow range of focal length
c) having a wide range of curvature
d) having a narrow range of curvature
View Answer

Answer: b
Explanation: Most lenses are made of spherical surfaces because spherical surfaces are easy to make and have wide range of curvature. Hence, the spherical surface of lens only results in having a wide range of curvature, not wide range of focal length.

3. What are Concave lenses?
a) Thicker from the centre than at the edge
b) Thinner from the centre than at the edge
c) Thicker from both the positions
d) Thinner from both the positions
View Answer

Answer: b
Explanation: Concave lenses are thinner from the centre than at the edge. However, convex lenses are thicker from the centre than at the edge.
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4. Out of the following which one is diverging or negative lens?
a) Biconvex
b) Plano-convex
c) Biconcave
d) Positive meniscus
View Answer

Answer: c
Explanation: Biconcave is the diverging or negative lens. Whereas biconvex, plano-convex and positive meniscus are converging or positive lenses.

5. What is the reciprocal, of the length of radius of curvature?
a) Focal length
b) Curvature
c) Optical centre
d) Power
View Answer

Answer: b
Explanation: The reciprocal of length of radius of curvature is called curvature.
i.e., C=1/R
while, the reciprocal of focal length is power.
i.e., P=1/f
A lens has two centres of curvature, also it has two radii of curvature, one for each of the refracting surface.
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6. Where does the principal axis intersect the two refracting surfaces?
a) Front vertex and back vertex
b) Focal length
c) Radius of curvature
d) Principal focus
View Answer

Answer: a
Explanation: The point at which the principal axis intersects, the two refracting surfaces are front vertex and back vertex. Focal length is the distance of focal point and optical centre. The length of radius of curvature of the surface, is called radius of curvature. The point at which converging and diverging of the rays take place is called principal focus.

7. What is the power of lens?
a) Reciprocal of R
b) Reciprocal of C
c) Reciprocal of f
d) Equal to f
View Answer

Answer: c
Explanation: The reciprocal of focal length is power.
i.e., P=1/f
while, the reciprocal of length of radius of curvature is called curvature.
i.e., C=1/R.
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8. If the object is at infinity, where and how will the image form?
a) At 2F, real, inverted and of same size
b) At F, point size, real and inverted
c) Between F and 2F, real, inverted and diminished
d) Beyond 2F, real, inverted and magnified
View Answer

Answer: b
Explanation: When the object is at infinity, the image will form at F. It will be of point size, real and inverted. Whereas, if the object is at 2F, the image will form at 2F and the image will be real, inverted and of the same size. Moreover, if the object is beyond 2F, the image will form, in between F and 2F, it will be real, inverted and diminished. While, if the object is in between F and 2F, the image will form beyond 2F, it will be real, inverted and magnified.

9. If the lens is placed in water, then the refractive index with respect to water will be ________
a) m=a/b
b) m=b/a
c) m*a=b
d) m/b=a
View Answer

Answer: a
Explanation: The refractive index of glass with respect to water, “m” is:
m=a/b
where, “m” is the refractive index of glass with respect to water, “a” is the refractive index of glass and “b” is the refractive index of water.
whereas, the refractive index of water with respect to glass, “n” (say) is:
n=b/a
=> n*a=b
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10. How is Lateral or Transverse magnification given?
a) m = h1/h2
b) m = h2/h1
c) m = h1*h2
d) m = h2 + h1
View Answer

Answer: a
Explanation: Lateral or Transverse magnification is “m” = h2/h1.
where, h1 is the length of object and h2 is the length of image. “m” is the lateral or transverse magnification.
It is the ratio of the length of image to the object size, both the lengths are measured perpendicular to the principal axis.

11. Name the bending of light, if the beam of light passes obliquely from one medium to another.
a) Refraction
b) Dispersion
c) Reflection
d) Deviation
View Answer

Answer: a
Explanation: If the beam of light passes obliquely from one medium to another, then the bending of light is known as refraction. Whereas, reflection means change in direction of light, at an interface, between two different mediums, such that the light returns into the medium from where it had originated. Dispersion is the property by which spreading of light takes place according to its color, as it passes through an object. Deviation is the bending of a beam of light, away from a straight line by an angle called angle of deviation.

12. From which part of the lens, the ray of light passes without deviation?
a) Optical centre
b) Focus
c) Centre of curvature
d) Pole
View Answer

Answer: a
Explanation: Optical centre is the part of the lens, through which the ray of light passes without deviation. Focus is the point from where converging and diverging of beam of light takes place, when a narrow beam of light, parallel to principal axis and close to it, is incident on the surface of mirror.

13. When will the convex lens give a real image?
a) Beyond optical centre
b) Beyond focus
c) Beyond centre of curvature
d) Between focus and curvature
View Answer

Answer: b
Explanation: When the object is situated beyond Focus, the convex lens will always give a real image. If the object is situated in between Focus and curvature, the image will be beyond 2F and the image will be virtual.

Sanfoundry Global Education & Learning Series – Engineering Physics.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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