CBSE 10th Standard Science Subject Light Reflection and Refraction Ncert Exemplar 5 Mark Questions 2021
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CBSE 10th Standard Science Subject Light Reflection and Refraction Ncert Exemplar 5 Mark Questions 2021
10th Standard CBSE
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Reg.No. :
Science
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Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.
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Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3 rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
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The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?
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Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of - 50 cm. What is the nature of the lens and its power used by each of them?
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(a) Define focal length of a divergent lens.
(b) A divergent lens of focal length 30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed.
(c) Draw a ray diagram to show the formation of image in the above situation.
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CBSE 10th Standard Science Subject Light Reflection and Refraction Ncert Exemplar 5 Mark Questions 2021 Answer Keys
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The following are the 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 angle of refraction is a constant, for the light of a given pair of media.This law is also known as Snell's law of refraction.The ray diagram is as shown.As seen in the refracted ray are in the same plane.
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Given m = 1/3, f = 20 cm, u = ?
Using the expression m = -v/u
We have 1/3 = -v/u or u = -3 V
Now using the expression
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
We have
\(\frac{1}{20}=\frac{1}{u}+\frac{-3}{u}=\frac{-2}{u}\)
or u = -40 cm -
Given m = -3, v = 80 cm, u = ?
Using the expression \(m=\frac { -v }{ u } \)
we have \(-3=\frac { 80 }{ u } \) or \(u=\frac { -80 }{ 3 } =-26.67cm\)
The image is real and the lens is a convex lens. -
It is defined as the ability of a lens to bend the rays of light it is given by the reciprocal of focal length in metre. Its unit is dioptre. The lens of focal length + 50 cm is a convex lens and the lens of focal length - 50 cm is a concave lens.
Their powers are
P1 = 1/F1 = 1/+0.5 = 2D
P2 = 1/F2 = 1/-0.5 = -2D -
(a) (i) The principal focus of a diverging lens is a point on its principal axis from which light rays, originally parallel to the axis appear to diverge after passing through the diverging (concave) lens.
(ii) The distance between pole and principal focus of a diverging lens is called the focal length in the given diagram OF is the focal length.
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(b) Divergent lens (Concave lens) Focal length, f = -30 cm
Size of the object, h1 = 6 cm
Image distance, v = -15 cm
Object distance, u = ?
Size of the image, h2 = ?
According to the lens formula,
\(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } \)
\(\frac { 1 }{ -30 } =\frac { -1 }{ -15 } +\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ u } =\frac { -1 }{ 15 } +\frac { 1 }{ 30 } =\frac { -2+1 }{ 30 } =\frac { -1 }{ 30 } \)
u=-30cm
\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } =\frac { v }{ u } \)
\(\Rightarrow \frac { { h }_{ 2 } }{ 6 } =\frac { -15 }{ -30 } \)
\(\Rightarrow { h }_{ 2 }=\frac { 15 }{ 30 } \times 6=+3cm\)
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