CBSE Class 10 Physics Light Reflection And Refraction Worksheet Set 05

This CBSE Class 10 Physics worksheet for Light – Reflection and Refraction (Chapter 9) gives 24 short practice questions with answers and no multiple choice questions. Students complete fill in the blanks, decide on true or false statements and solve numericals on convex and concave mirrors, convex lenses and lens power, with each answer shown step by step.

Fill in the Blanks

Question 1. Fill in the blank(s): The distance between ________________ and ________________ of the lens is known as the focal length of a lens.
Answer:
The distance between the optical centre and the principal focus of the lens is known as the focal length of the lens.
In simple words: Focal length is measured from the optical centre to the focus.

Teacher's Note:
1. Focal length is positive for a convex lens and negative for a concave lens.
2. It is measured along the principal axis.

Question 2. Fill in the blank(s): The refractive index of any transparent medium is the ratio of the speed of light in ________________ to that in the ________________ .
Answer:
The refractive index is the ratio of the speed of light in vacuum to that in the medium: n = c/v.
In simple words: n = speed in vacuum divided by speed in the medium.

Teacher's Note:
1. The value is always greater than 1.
2. It has no unit.

Question 3. Fill in the blank(s): The image is ________________ and ________________ if the magnification has a minus sign.
Answer:
The image is real and inverted if the magnification has a minus sign.
In simple words: Negative m means a real, upside-down image.

Teacher's Note:
1. Positive m means a virtual, erect image.
2. This sign rule holds for both mirrors and lenses.

Question 4. Fill in the blank(s): ________________ mirror is used as a security mirror on roads at sharp bends and concealed entrances.
Answer:
A convex mirror is used as a security mirror at sharp bends and concealed entrances, because it gives a wide field of view.
In simple words: A convex mirror shows a wide view of the road.

Teacher's Note:
1. It also gives an erect, diminished image.
2. Convex mirrors are used on vehicles for the same reason.

Question 5. Fill in the blank(s): For calculating the power of a lens, we need its focal length in ________________ .
Answer:
The focal length must be in metres, so that the power comes out in dioptres: P = 1/f (m).
In simple words: Use metres for focal length when finding power in dioptres.

Teacher's Note:
1. Convert centimetres to metres by dividing by 100.
2. 1 D = 1 m⁻¹.

Question 6. Fill in the blank(s): The focal length of a concave lens is considered to be ________________ .
Answer:
The focal length of a concave lens is considered to be negative as per the Cartesian sign convention.
In simple words: A concave lens has a negative focal length.

Teacher's Note:
1. Its power is also negative.
2. A convex lens has a positive focal length.

Question 7. Fill in the blank(s): In Vacuum the speed of light is ________________.
Answer:
In vacuum the speed of light is 3 × 10⁸ m/s (the maximum possible speed).
In simple words: Light moves at 3 × 10⁸ m/s in vacuum.

Teacher's Note:
1. The speed is a little smaller in air and much smaller in glass.
2. Use this value in v = c/n.

Question 8. Fill in the blank(s): At whatever distance you stand from a spherical mirror, your image appears erect. Then that mirror may be ________________ .
Answer:
The mirror may be a plane or convex mirror.
A concave mirror gives an inverted image when you stand beyond its focus.
In simple words: Always-upright images come from plane or convex mirrors.

Teacher's Note:
1. The convex mirror image is diminished and the plane mirror image is of the same size.
2. Think about which mirrors form only virtual images.

Question 9. Fill in the blank(s): A positive sign of magnification indicates that the image is ________________ .
Answer:
A positive sign of magnification indicates that the image is virtual (and erect).
In simple words: Positive m means a virtual, erect image.

Teacher's Note:
1. Real images are always inverted and have a negative m.
2. A plane mirror has m = +1.

Question 10. Fill in the blank(s): A ________________ lens will always give a virtual, erect and diminished image, whatever be the position of the object.
Answer:
A concave lens always gives a virtual, erect and diminished image.
In simple words: Only a concave lens always shrinks the image.

Teacher's Note:
1. A convex lens can also give real or enlarged images.
2. The image lies between the lens and its focus.

True or False

Question 11. State whether true or false: Lateral displacement increases with the thickness of the given slab.
Answer:
True. A thicker slab gives a larger sideways shift of the emergent ray (it also increases with the angle of incidence).
In simple words: A thicker glass slab shifts the ray more.

Teacher's Note:
1. The emergent ray stays parallel to the incident ray.
2. The shift is zero for normal incidence.

Question 12. State whether true or false: A person standing in front of a mirror finds his image larger than himself. This nature of image shows that mirror is convex in nature.
Answer:
False. An enlarged image is formed by a concave mirror (when the person is within its focal length). A convex mirror always gives a diminished image.
In simple words: A magnified image means a concave mirror, not a convex one.

Teacher's Note:
1. Convex mirrors only reduce the size of the image.
2. The person must stand between the pole and the focus.

Question 13. State whether true or false: A convex lens of refractive index = 3/2 is placed in water of refractive index 4/3, the focal length of the lens will increase inside water.
Answer:
True. The glass lens is optically less dense relative to water than relative to air (1.5/1.33 instead of 1.5), so it bends light less and its focal length increases.
In simple words: A glass lens is weaker in water, so its focal length is longer.

Teacher's Note:
1. A lens bends light because of the difference in refractive index with its surroundings.
2. If the medium had the same refractive index as the lens, the focal length would be infinite.

Question 14. State whether true or false: A convex mirror and a concave lens diverge the rays falling parallel to the principal axis.
Answer:
True. Both are diverging devices. After reflection or refraction the parallel rays spread out and appear to come from the focus behind the mirror or on the object side of the lens.
In simple words: Both a convex mirror and a concave lens spread parallel rays out.

Teacher's Note:
1. Each has a virtual focus.
2. Concave mirror and convex lens are the converging pair.

Question 15. State whether true or false: The power of a concave lens must be positive.
Answer:
False. A concave lens has a negative focal length, so its power P = 1/f is negative.
In simple words: A concave lens has negative power.

Teacher's Note:
1. Only convex lenses have positive power.
2. The unit remains the dioptre.

Numericals

Question 16. An object placed at distance from a convex lens of focal length 25 cm such that the size of the images is 1/4th of the object. What will be the object distance?
Answer:
A convex lens gives a diminished image only when it is real, so m = −1/4 and v = −u/4 in magnitude: |v| = |u|/4.
\( \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{4}{|u|}+\dfrac{1}{|u|}=\dfrac{5}{|u|} \)
So |u| = 5f = 5 × 25 = 125 cm. The object is 125 cm in front of the lens (and the image is 31.25 cm behind it).
In simple words: For a real image 1/4 the size, the object is 5 focal lengths away: 125 cm.

Teacher's Note:
1. A virtual image from a convex lens is always enlarged, so it cannot be one-fourth.
2. Check: v = 31.25 cm and v/u = 1/4.

Question 17. If magnification of a body of size 2 m is 4, then what will be the size of the image (in metres)?
Answer:
m = h′/h, so h′ = m × h = 4 × 2 = 8 m.
In simple words: Multiply the object size by the magnification: 8 m.

Teacher's Note:
1. The image is four times as big as the object.
2. A positive m means the image is erect.

Question 18. What will be the power of a lens with focal length 3 cm?
Answer:
f = 3 cm = 0.03 m (taking the lens as convex, f positive).
P = 1/f = 1/0.03 = +33.3 D.
In simple words: Convert 3 cm to 0.03 m and take the reciprocal: about 33.3 D.

Teacher's Note:
1. Always change centimetres to metres before using P = 1/f.
2. A very short focal length gives a very high power.

Question 19. A 1.2 cm long pin is placed perpendicular to the principal axis of a convex mirror of focal length 12 cm, at a distance of 8 cm from the mirror. (a) Find the location of the image. (b) Find the height of the image. (c) Is the image erect or inverted?
Answer:
f = +12 cm, u = −8 cm, h = 1.2 cm.
(a) \( \dfrac{1}{v}=\dfrac{1}{12}+\dfrac{1}{8}=\dfrac{5}{24} \), so v = +4.8 cm (4.8 cm behind the mirror).
(b) m = −v/u = −4.8/(−8) = +0.6, so h′ = 0.6 × 1.2 = 0.72 cm.
(c) h′ is positive, so the image is erect (virtual and diminished).
In simple words: The image is 4.8 cm behind the mirror, upright and 0.72 cm tall.

Teacher's Note:
1. For a convex mirror the image is always behind it.
2. Positive h′ means an erect image.

Question 20. An object 4 cm high is placed 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed to obtain a sharp image? Find the nature and size of the image.
Answer:
f = −15 cm, u = −25 cm, h = 4 cm.
\( \dfrac{1}{v}=-\dfrac{1}{15}+\dfrac{1}{25}=-\dfrac{2}{75} \), so v = −37.5 cm. The screen must be 37.5 cm in front of the mirror.
m = −v/u = −(−37.5)/(−25) = −1.5, so h′ = −1.5 × 4 = −6 cm.
The image is real, inverted and enlarged, 6 cm tall.
In simple words: Put the screen 37.5 cm from the mirror; the image is upside down and 6 cm tall.

Teacher's Note:
1. The object lies between F and C, so the image is beyond C.
2. A screen can only receive a real image.

Question 21. An object of height 2.0 cm is placed perpendicular to the principal axis of a concave mirror. The distance of the object from the mirror is 30 cm and its image is formed 60 cm from the mirror on the same side of the mirror as the object. Find the height of the image.
Answer:
The image is on the same side as the object, so it is real: u = −30 cm, v = −60 cm.
m = −v/u = −(−60)/(−30) = −2.
h′ = m × h = −2 × 2.0 = −4.0 cm.
The image is 4.0 cm tall and inverted. (The mirror formula gives f = −20 cm.)
In simple words: The image is twice as tall as the object, 4 cm, and inverted.

Teacher's Note:
1. Image and object on the same side of a mirror means a real image.
2. A negative height means inverted.

Question 22. The power of a lens is 2.5 dioptre. What is the focal length and the type of lens?
Answer:
f = 1/P = 1/2.5 = 0.4 m = +40 cm.
The power (and focal length) is positive, so it is a convex (converging) lens.
In simple words: A power of +2.5 D gives f = 40 cm, a convex lens.

Teacher's Note:
1. Positive power means a convex lens.
2. The focal length comes out in metres.

Question 23. A convex lens has a focal length of 20 cm. Calculate at what distance from the lens should an object be placed so that it forms an image at a distance of 40 cm on the other side of the lens. State the nature of the image formed?
Answer:
f = +20 cm, v = +40 cm (image on the other side).
\( \dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=\dfrac{1}{40}-\dfrac{1}{20}=-\dfrac{1}{40} \), so u = −40 cm.
The object is at 2f (40 cm). The image is real, inverted and of the same size (m = −1).
In simple words: The object has to be 40 cm from the lens, and the image is real, upside down and the same size.

Teacher's Note:
1. Object and image at equal distances means the object is at 2f.
2. m = v/u = −1.

Question 24. A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30 cm. The distance of the object from the lens is 20 cm. Find the (i) position (ii) nature and (iii) size of the image formed.
Answer:
f = +30 cm, u = −20 cm, h = 10 cm.
\( \dfrac{1}{v}=\dfrac{1}{30}-\dfrac{1}{20}=-\dfrac{1}{60} \), so (i) v = −60 cm: the image is 60 cm from the lens on the same side as the object.
m = v/u = (−60)/(−20) = +3, so (iii) h′ = 3 × 10 = 30 cm.
(ii) The image is virtual, erect and enlarged.
In simple words: The object lies inside the focus, so the lens acts as a magnifier: a 30 cm tall virtual image 60 cm away.

Teacher's Note:
1. u = 20 cm is less than f = 30 cm, so expect a virtual image.
2. The distance is measured from the optical centre of the lens.