Reflection and Refraction of Light Important questions
QUESTIONS: (5 marks each) Question 1. With the help of ray diagrams, show the formation of images of an object by a concave mirror. When it is placed
(i)beyond the centre of curvature
(ii)at the centre of curvature.
Question 2. Draw ray diagrams to show the formation of image by an object by a concave lens when the object is placed
(i)at infinity
(ii)between infinity and optical centre of the lens.
Question 3. Draw a labelled ray diagram to locate the image of an object formed by a convex lens of Focal length 20 cm when the object is placed 30 cm away from the lens.
Question 4. What is meant by the power of a lens? What is its SI unit? Name the type of lens whose power is positive. The image of an object formed by a lens is real, inverted and of the same size as the object. If the image is at a distance of 40 cm from the lens, what is the nature and power of the lens? Draw ray diagram to justify your answer. Answers
(i) Power of lens is the ability of a lens to converge or diverge light rays and it is equal to Reciprocal of focal length of the lens
(ii) S.I. unit is dioptre
(iii) Convex lens has positive power
(iv) v= +40 cm; h' = h The lens is real, inverted and same sized
u =-v
Using lens Formula,
$\frac {1}{f} = \frac {1}{v} -\frac {1}{u}$
2f=v or f=20 cm
It is a convex lens
Question 5. One half of a convex lens of focal length 10 cm is covered with a black paper. Can such a lens produce an image of a complete object placed at a distance of 30 cm from the lens? Draw a ray diagram to justify your answer.
Question 6. With the help of a ray diagram, state and explain the laws of reflection of light at a plane mirror. Mark the angle of incidence and reflection clearly in the diagram. If the angle of incidence is 32.5o, what will be the angle of reflection?
Question 7. It is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm.
(i)What should be the range of distance of an object placed in front of the mirror?
(ii)Will the image be smaller or larger than the object? Draw ray diagram to show the formation of image in this case,.
(iii)Where will the image of this object be, if it is placed 6 cm in front of the mirror? Draw ray diagram for this situation to justify your answer.
Show the position of pole, principal focus and the centre of curvature in the ray diagram. Answers
(i) In concave mirror erect is formed only when the object is placed between pole and focus. So object distance should be less than 12 cm
(ii) The image would be larger than the object
(iii) Here u=-6 cm , f=-12 cm
Using Mirror Formula
$\frac {1}{f} =\frac {1}{v} + \frac {1}{u}$
$\frac {1}{-12} =\frac {1}{v} + \frac {1}{-6}$
v= 12 cm
Question 8. A student has three concave mirrors A, B and C of focal lengths 20 cm, 15 cm and 10 cm respectively. For each concave mirror he performs the experiment of image formation for three values of object distance of 30 cm, 10 cm and 20 cm.
Giving reason answer the following:
(a)For the three object distances, identify the mirror which will form an image equal in size to that of object. Find at least one value of object distance.
(b)Out of the three mirrors, identify the mirror which would be preferred to be used for shaving purpose.
(c)For the mirror B, draw ray diagram for image formation for any two given values of object distance. Answers
Given
$f_A=20 cm$,$f_B=15 cm$,$f_C=10 cm$,$u_1=30 cm$,$u_2=10 cm$,$u_3=20 cm$
a. For same size image, object should be placed at Center of Curvature
So for Mirror A, Image will not of same size for these positions
For Mirror B,Image will be of same size for position $u_1=30 cm$
For mirror C,Image will be of same size for position $u_3=20 cm$
b.We need enlarged and erect image for shaving. For erect and enlarged image, Object should be placed between Pole and Focus. Face would be generally kept at more than 10 cm distance, so mirrors A and B are suitable for shaving
Question 9. An object is placed at infinity from a concave mirror of focal length 10 cm. Find the position and nature of image formed. Draw a ray diagram to show the formation of image. (Not to scale) Answers
Image formed at 10 cm in front of concave mirror,inverted,smaller than object.
Question 10. (a) Define focal length of a spherical lens.
(b) A divergent lens has a focal length of 30 cm. At what distance should an object of height 5 cm from the optical centre of the lens be placed so that its image is formed 15 cm away from the lens? Find the size of the image also.
(c) Draw a ray diagram to show the formation of image in the above situation. Answers
(a) check this Refraction by Spherical Lenses
(b) f= -30 cm; $h_o = 5cm$ ,v=-15 cm
Using lens Formula,
$\frac {1}{f} = \frac {1}{v} -\frac {1}{u}$
$\frac {1}{-30} = \frac {1}{-15} -\frac {1}{u}$
u=-30 cm
Now
$ \frac {v}{u}=\frac {h_i}{h_o}$
or
$h_i = \frac {v \times h_o}{u} = 2.5 cm$
Question 11. (a) A converging lens forms a real and inverted image of an object at a distance of 100 cm from it. Where should an object be placed in front of the lens, so that the size of the image is twice the size of the object? Also, calculate the power of a lens.
(b) State laws of refraction. Answers
a. Since this is converging lens, it is a convex lens
v=+100 cm
Now Magnification for lens is given by
$m= \frac {v}{u} =\frac {h_i}{h_o}$
Since the image is inverted
$\frac {v}{u} =-2$
or u =-50 cm
The object is placed at a distance 50 cm from the lens
Now using Lens formula
$\frac {1}{f} = \frac {1}{v} -\frac {1}{u}$
$\frac {1}{f} = \frac {1}{100} -\frac {1}{-50}$
or f=33.33 cm
Now
Power of the lens
$power = \frac {1}{f(m} = 3D$
Question 12. List the new Cartesian sign convention for reflection of light by spherical mirrors. Draw a diagram and apply these conventions for calculating the focal length and nature of a spherical mirror which forms a 1/3 times magnified virtual image of an object placed 18 cm in front of it. Answers
a. check this Sign convention for reflection by spherical mirrors
b. Here u=-18 cm, m=1/3
Now $m= -\frac {v}{u}$
or v = 6 cm
Now using Mirror Formula
$\frac {1}{f} = \frac {1}{v} +\frac {1}{u}$
$\frac {1}{f} = \frac {1}{6} +\frac {1}{-18}$
f=9 cm.
Hence a convex mirror
Question 13.
a.One – half of a convex lens of focal length 110 cm is covered with a black paper. Can such a lens produce an image of a complete object placed at a distance of 30 cm from the lens? Draw a ray diagram to justify your answer.
b. A 4 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 15 cm. Find the nature, position and size of the image. Answers
a. Yes
b. u=-15 ,f=20 cm ,v=?
Now using Lens formula
$\frac {1}{f} = \frac {1}{v} -\frac {1}{u}$
$\frac {1}{20} = \frac {1}{v} -\frac {1}{-15}$
v=-60 cm
Now $m= \frac {v}{u}=\frac {h_i}{h_o}$
or $h_i=16 cm$
So image is enlarged and erect
Question 14. Why does a light ray incident on a rectangular glass slab immersed in any medium emerge parallel to itself? Explain using a diagram.
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