Light - Reflection and refraction
1. Which one of the following materials cannot be used to make a lens?
a) Water b) Glass c) Plastic. d) Clay
Explanation :- Because clay is an opaque substance and it doesn't allow the light to pass through it.
2. The image formed by a concave mirror is observed to be virtual ,erect and
larger than the object .Where should be the position of the object?
a) Between the principal focus and the centre of curvature
b) At the centre of curvature
c) Beyond the centre of curvature
d) Between the pole of the mirror and it's principal focus.
Explanation :- one and only condition that forms a virtual image using concave mirror is, the position of the object placed between pole and focus
3. Where should an object be placed in front of a convex lens to get a real
image of the size of the object?
a) At the principal focus of the lens
b) At twice the focal length
c) At infinity
d) Between the optical centre of the lens and it's principal focus.
Explanation :- if we place an object at center of curvature then the image also forms at center of curvature with same size, real and inverted.
4. A mirror and a thin spherical lens have each a focal length of -15 cm.The
mirror and the lens are likely to be
a) both Concave.
b) both convex.
c) the mirror is concave and the lens is convex.
d) the mirror is convex, but the lens is concave.
Explanation :- All the distances measured to the left of the origin (along – x-axis) are taken as negative.
In case of concave mirror and Concave lens we measure focal length in left side direction
5. No matter how far you stand from a mirror, your image appears erect.The mirror is likely to be
a) only plane. b) only concave.c) only convex.
d) either plane or convex.
Explanation :- plane and convex mirrors always form virtual, erect images
6. Which of the following lenses would you prefer to use while reading small
letters found in a dictionary?
a) A convex lens of focal length 50 cm.
b) A concave lens of focal length 50 cm.
c) A convex lens of focal length 5 cm.
d) A concave lens of focal length 5 cm.
Explanation :- Magnification is more for convex lenses with shorter focal length.
7. We wish to obtain an erect image of an object, using a concave mirror of
focal length 15 cm. What should be the range of distance of the object from
the mirror?
What is the nature of the image?
Is the image larger or smaller than the object? Draw a ray diagram to show
the image formation in this case.
1. We can obtain a virtual and erect image by a concave mirror , only when the object is placed in between the principal focus and the pole of the mirror.
2. Therefore the range of distance of the object from the mirror is 0 to 15 cm.
3. Nature of the image: virtual, erect and enlarged.
4. Image is larger than the object.
8. Name the type of mirror used in the following situations.
a) Headlights of a car.
b) Side/rear - view mirror of a vehicle.
c) Solar furnace.
Support your answer with reason.
a) concave mirror.
Reason:- The parallel beam of light helps us to see things at a distance. we can obtain a parallel beam of light only when the bulb is placed at focus of a concave mirror. So we use a concave mirror in the headlights of a car.
b) convex mirror.
Reason :- Convex mirror not only can form an erect and virtual image but also forms a diminished image which helps the driver to see a larger area behind the vehicle. So, Convex mirrors are used as rear-view mirrors
c) concave mirror.
Reason :- The parallel beam of light from the sun converges to a point (focus) along with the heat when it incident on a concave mirror. That point (focus ) will be very hot. That is why we use concave mirrors in Solar furnaces.
9. One- half of a convex lens is covered with a black paper.Will this lens produce a complete image of the object ? Verify your answer experimentally .Explain your observations.
Yes, the lens will still produce a complete image of the object, even with one-half of it covered with black paper.
Reason :- A convex lens converges light rays, and the image formation depends on the convergence of light rays, not on the entire lens surface.
With one-half of the lens covered, the remaining half will still converge light rays and form an image.
Experimental Verification:-
Procedure :-
1. Take a candle,a bi convex lens - half of it covered half with black paper, lens holder and a screen.
2. Place the lens on a lens holder.
3. Place an object (candle ) in front of the lens at a distance.
4. Adjust the position of the screen until an image of the candle is observed on it.
5. Observe the projected image on the screen.
6. Repeat this activity by removing black paper from the lens and observe the image formed on the screen.
7. Compare the images formed in both situations.
Observations:-
1.In both situations You will see a complete image on screen.
2. But the image formed by a half -covered lens is slightly dimmer due to reduced light intensity.
Result :-
This demonstrates that even with half the lens covered, the convex lens can still produce a complete image of the object.
However, the image quality and brightness may be affected.
10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10
cm. Draw the ray diagram and find the position , size and the nature of the image formed.
f = 10cm
ho = 5cm
Image position v =?
Lens formula,$$ \frac 1f = \frac 1 v – \frac 1 u$$
$$\frac{1} {10} = \frac 1 v – \frac{1} {-25} $$
$$\frac{1} {10} = \frac1v + \frac{1} {25} $$
$$\frac1v = \frac1{10} – \frac1{25} $$
$$\frac1v = \frac{5 –2} { 50} $$
$$\frac 1v = \frac{3} {50} $$
$$ v = \frac{1} {\frac{50} {3}} $$
$$v = \frac{50} {3} = 16.67cm.$$
The image will be formed at 16.67cm from the lens .
iii) Size of the image (hi) :5cm
Since $$\frac{h_i} {h_o} = \frac{v} {u} $$
$$\frac {h_i} {5} =\frac {50}{3×-25} $$ $$\frac {h_i} {5} =\frac {2}{-3}$$
$$ {h_i} =\frac {2×5}{-3}$$
$$ {h_i} =\frac {-10}{3}$$
= – 3.33 cm
iv) Magnification, $$ {m} =\frac {v}{u}$$
$$=\frac{\frac{50} {3} }{-25}$$
$$= \frac{-50}{(3×25)}$$
$$\frac{ -2} {3}$$
= -0.67
Since, h i < ho
Therefore, Image is smaller than the object. Negative sign indicates the image is real and inverted.
11. 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.
Given, f = -15 cm
v = -10 cm
u = ?
Lens formula, $$\frac{1} {f} =\frac {1} {v} - \frac{1} {u} $$
$$\frac{1} {u} =\frac {1} {v} - \frac{1} {f} $$
$$= \frac{1}{-10} – \frac{1} {-15} $$
$$=\frac{-1} {10} + \frac{1} {15} $$
$$\frac{1} {u} = \frac{(-3+2)} {30} $$
$$\frac{1} {u} = \frac{-1}{30}$$
:. u = (-30) cm
Nature of the image is virtual, erect, diminished
12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image?
Given:- u = -10 cm
f = 15 cm
v = ?
$$Mirror formula, \frac1f = \frac1u + \frac1v$$
$$\frac1 v = \frac1f - \frac1u$$
$$\frac1v = \frac{1} {15} -\frac{1} {-10} $$
$$\frac1 v = \frac{1} {15} + \frac{1} {10} $$
$$= \frac{2+3} { 30} $$
$$\frac1v = \frac{5} {30} $$
$$\frac 1v =\frac16$$
:. Image distance,v = 6 cm
$$Since, magnification, m =\frac{-v}{u} $$
$$M = \frac{-6} {-10} $$
M= + 0.6
Nature of the image:-
i) diminished image
ii) positive sign indicates it is an erect and virtual image.
13. The magnification produced by a plane mirror is + 1 . What does this mean?
Given, Magnification = +1
Then Size of image / size of object = 1
:. Size of Image = size of object
Also the magnification is positive.
Hence the image formed is virtual and erect.
14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Given, ho = 5cm
u = -20 cm
R = +30 cm
f = +15 cm
$$mirror formula,\frac1 f = \frac1v + \frac1u$$
$$\frac1v = \frac1f – \frac1u$$
$$= \frac{1} {15} – \frac{1} {-20} $$
$$=\frac{1} {15} + \frac{1} {20} $$
$$= \frac{4+3} { 60} $$
$$\frac1v = \frac{7} {60} $$
$$v = \frac1{\frac{60} {7} } $$
$$:. v = \frac{60} {7} $$
v = 8.57 cm
$$Magnification, m = \frac{h_i} {h_o} =\frac {-v} {u} $$
$${h_i} =\frac{- v} {u} × {h_o} $$
$$= \frac{- 60} {7} × \frac{5} {-20} $$
$$= \frac{300} {140} $$
$$= \frac{15} {7} $$
= 2.14 cm
:. Height of the image = 2.14 cm
Since. hi < ho
Nature of the image:-
Diminished, erected and virtual image.
15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed , so that a sharp focused image can be obtained ? Find the size and nature of the image.
Given, ho= 7 cm
u = -27 cm
f = -18 cm
v = ?
Mirror formula,
$$\frac1f = \frac1v +\frac1u $$
$$\frac1v = \frac1f - \frac1u $$
$$= \frac{1} {-18} – \frac{1} {-27} $$
$$= \frac{-1} {18} + \frac{1} {27} $$
$$=\frac{-3+2} {54} $$
$$\frac1v = \frac{-1} {54} $$
:. v = -54 cm.
so, the screen should be placed at 54 cm from the mirror.
Magnification,$$m =\frac{ h_i} {h_o} = \frac{-v} {u} $$
$$hi = \frac{- v} {u} × ho$$
$$=\frac {- (- 54)} {- 27} × 7$$
= - 14cm
:. height of the object = -14 cm.
Since hi > ho
So, the nature of the image is virtual, erect and magnified.
16. Find the focal length of a lens of power –2.0 D. What type of lens is this?
Given, p = –2 D
$$f = \frac1p$$
$$= \frac{1} {–2} $$
= –0.5 m
:. f = – 50 cm
Negative sign indicates, the given lens is concave lens.
17. 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?
Given, p = +1.5 D
$$f = \frac1p$$
$$= \frac{1} {1.5} =\frac{10}{15} $$
$$= \frac23$$
:. f = +0.67 m.
Positive sign indicates, the given lens is converging (convex) lens
1. Find the focal length of a convex mirror whose radius of curvature is 32 cm?
Given, radius of curvature (R) = 32 cm
Relation between focal length f and radius of curvature R is $$f = \fracR 2$$
$$= \frac{32} {2} $$
= 16 cm.
So, focal length is 16cm
2. A concave mirror produces a three times magnified (enlarged ) real image of an object placed at 10 cm in front of it. Where is the image located?
Given, Object distance (u) = -10 cm
magnification, (m ) = - 3
Image distance,v = ?
Since, $$m = \frac{–v} {u}
=> $$- 3 = \frac{-v} {-10} $$
$$- 3 = \frac{v} {10} $$
v = - 30 cm
So, the object distance is 30cm
1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? why?
1. Air is optically rarer medium and water is denser medium.
2."When a light ray travels from a rarer medium (air) to a denser medium (water), due to the decrease in speed, it bends towards the normal."
3.This phenomenon is known as refraction.
2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vacuum is 3×108 m/s.
Given :- n = 1.5
c = 3×108 m/s
v = ?
Formula for refractive index $$n = \fraccv$$
=> $$v = \frac c n$$
$$= \frac{3×10^8} {1.5} $$
v = 2×108 m/s.
So, Speed of light in glass = 2×108 m/s.
3. Find out from table 9.3,the medium having highest optical density. Also find the medium with lowest optical density.
The medium having the highest optical density is diamond.
The medium having the lowest optical density is air.
4. You are given kerosene, turpentine and water in which of these does the light travel fastest? ( use the information given in table 9.3 page number 32 of the textbook) .
1. The refractive indices of kerosene, turpentine, and water are 1.44, 1.47, and 1.33, respectively.
2. Since light travels faster in the medium with the lower refractive index, it travels fastest in water among the given media.
5. The refractive index of diamond is 2.42. What is the meaning of this statement?
1. The refractive index compares the speed of light in a medium with respect to a vacuum.
2. The refractive index of diamond is 2.42, which means that light travels about 2.42 times slower through diamond than through a vacuum.
1. Define 1 diopter of power of lens?
1 dioptre is the power of a lens whose focal length is 1 metre. 1D = 1m - 1
2. A Convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens If the size of the image is equal to size of the object. Also find the power of the lens?
1. If the object is placed at the center of curvature (2F) of the convex lens, then the image characteristics are:
- Position: at center of curvature (2F)
- Nature: real and inverted
- Size: same size as the object
2. To form an image of the same size as the object, which is real and inverted, and positioned 50 cm (at C) from the lens, the object must also be placed at 50 cm (at C) from the lens. This means the object is placed at the center of curvature (2F) of the convex lens.
3. So, u = -50cm,v = 50cm, R = 50cm, m=-1
4. $$f = \frac R 2 = \frac{50} {2} = 25 cm$$
5. Power of lens,$$ p = \frac{100} {f} , (f in cm)$$
$$= \frac{100} {25} $$
:. p = +4 D
3. Find the power of a concave lens of focal length 2 m.
Given, f = -2m (as per sign convention)
Since,$$ p = \frac 1 f, (f in metres)$$
$$=\frac{ 1} {-2} $$
p = – (0. 5) D
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