CBSE Class 10 Physics Reflection and Refraction of Light Worksheet Set A Solved

VERY SHORT ANSWER QUESTIONS

Question : If the image formed by a spherical mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a labelled ray
diagram to support your answer.
Answer : If the image formed by a spherical mirror is always erect and diminished then it is convex mirror.

Question : The linear magnification produced by a spherical mirror is -1. Analysing this value state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror. Draw any diagram to justify your answer.
Answer : (i) Concave mirror because the image is real, inverted.
(ii) Object is placed at C.

Question : State the laws of refraction of light. Explain the term absolute refractive index of a medium’ and write an expression to relate it with the speed of light in vacuum.
Answer : (a) Laws of refraction of light:
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
(ii) The ratio of sine of angle of incidence to the sine of the angle of refraction is constant, for the light of a given colour and for the given pair of media.
This law is also known as Snell’s law of refraction.
sini/sinr = constant,
where i is the angle of incidence and r is the angle of refraction.
This constant value is called refractive index of the second medium with respect to the first when the light travels from first medium to second medium.
⇒ constant = n21 = v1/v2 ∴sini/sinr = v1/v2
If n is the absolute refractive index of the medium, c is the velocity of light in vacuum and v is the speed of light in a given medium, then n = c/v.

Question : Draw ray diagrams to show the formation of three times magnified (a) real, and (b) virtual image of an object by a converging lens. Mark the positions of O, F and 2F in each diagram.
Answer :

(b) Ray diagrams of an object placed between F1 and optical centre O of lens can be drawn as follows:

(i) The image formed is virtual and erect.
(ii) Image is formed in front of the lens.
(iii) Image formed is enlarged.

Question : State the two laws of reflection of light.
Answer : Laws of reflection of light states that
(i) The angle of incidence is equal to the angle of reflection.
(ii) The incident ray, the reflected ray and the normal to the mirror at the point of incidence all lie in the same plane.

SHORT ANSWER TYPE QUESTIONS

Question : A concave mirror has a focal length of 20 cm. At what distance from the mirror should a 4 cm tall object be placed so that it forms an image at a distance of 30 cm from the
mirror? Also calculate the size of the image formed.
Answer : Given f = -20 cm v = -30 cm, u = ?
Using 1/v + 1/u = 1/f
1/u = 1/f – 1/v = 1/(−20) – /(−30) = (−3+2)/60
⇒ u = -60 cm
∴ Object placed at 60 cm from the mirror.
Also magnification, m = h′/h = −v/u
⇒ h’ = −(−30)/−60 × 4 = -2 cm
∴ The size of the image is 2 cm.

Question : The image of an object formed by a mirror is real, inverted and is of magnification -1.
If the image is at a distance of 40 cm from the mirror, where is the object placed? Where would the image be if the object is moved 20 cm towards the mirror? State reason and also draw ray diagram for the new position of the object to justify your answer.
Answer : Since the image formed by the mirror is real and inverted, therefore the mirror is concave and magnification of the mirror will be
m = –v/u ⇒ -1 = –v/u ⇒ v = u
i.e., object and image both are formed at the centre of curvature, i.e., 40 cm from the mirror.

Now, if the object is moved 20 cm towards the mirror, the object will be at the focus of the mirror and therefore the image will be formed at infinity.

Question : The refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. If speed of light in glass is 2 × 108 m/s, find the speed of light in water.
Answer : 

Question : (a) Water has refractive index 1.33 and alcohol has refractive index 1.36. Which of the two medium is optically denser? Give reason for your answer.
(b) Draw a ray diagram to show the path of a ray of light passing obliquely from water to alcohol.
(c) State the relationship between angle of incidence and angle of refraction in the above case.
Answer : (a) Here, alcohol is optically denser medium as its refractive index is higher than that of water. When we compare the two media, the one with larger refractive index is called the optically denser medium than the other as the speed of light is lower in this medium.
(b) Since light is travelling from water (rarer medium) to alcohol (denser medium), it slows down and bends towards the normal.

where i = angle of incidence and r = angle of refraction.
(c) According to Snell’s law,
sini/sinr=μalcohol /μwater =1.36/1.33 = 1.0225
∴ sin i = 1.0225 × sin r

Question : What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of -20 cm. Write the nature and power of each lens.
Answer : Power is the degree of convergence or divergence of light rays achieved by a lens.
It is defined as the reciprocal of its focal length.
i.e., P = 1/f
Given that: Focal length of lens A, fA = +40 cm
Focal length of lens B, fB = -20 cm
Lens A is converging. Lens B is diverging.

LONG ANSWER TYPE QUESTIONS

Question : Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose.
(i) State the nature of the lens and reason for its use.
(ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object?
(iii) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image.
Answer : (i) The lens used here is a convex lens and it is used as a magnifying glass because at close range, i.e., when the object is placed between optic centre and principal focus it
forms an enlarged, virtual and erect image of the object.
(ii) When this lens is placed such that the object is between the centre of curvature and the principal focus, the palmist obtain a real and magnified image.
(iii) Given focal length, f = 10 cm and u = -5 cm According to lens formula,

Thus, the image will be formed at 10 cm on the same side of the palm and the size of the image will be enlarged.

Question : (a) A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position, nature and size of the image formed.
(b) Draw a labelled ray diagram showing object distance, image distance and focal length in the above case.
Answer : (a) Given, h = 5 cm, f = 20 cm, u = -30 cm
Using lens formula, 1/v – 1/u = 1/f
1/v=1/u+1/f=1/(−30)+1/20=(−2+3)/60=1/60
⇒ v = 60 cm
Now, magnification, m = h′/h = v/u
⇒ h’ = v/u × h = 60/(−30) × 5 = -10 cm
Hence, the image formed at 60 cm, which is real and magnified.

Question : A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it”. Draw ray diagram to justify this statement stating the position of the object with respect to the lens in each case. An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.
Answer : Magnified erect image:

Given that h = 4 cm, u = -20 cm, f = -10 cm
Lens formula:
1/v – 1/u = 1/f ∴ 1/v – 1/(−20) = 1/(−10)
or 1/v=−1/10−1/20=(−2−1)/20=−3/20 or v = −20/3 cm

Question : Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
Answer :

Therefore, the mirror is concave and the image is real, inverted and diminished.

Question : (i) A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
(ii) A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer : (i) Power of lens (P) = 1/f
P = 1.5D
f = 1/1.5 = 10/15 = 0.66 m
A convex lens has a positive focal length. Therefore, it is a convex lens or a converging lens.
(ii) Focal length of concave lens (OF1), f = – 15 cm
Image distance, v= – 10 cm
According to the lens formula,

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