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Question (1):
Define refraction. |
| Answer:
The deviation in the path of light when it passes from one medium to another medium of different density is called refraction. |
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Question (2):
Define refractive index. |
| Answer:
The ratio of the speed of light in vacuum to the speed of light in a medium is called the refractive index of the medium. |
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Question (3):
What is the unit of refractive index? |
| Answer:
Refractive index is the ratio of velocity of light in two media and hence it is a mere number without any unit. |
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Question (4):
List out the factors on which the refractive index of a medium depends. |
Answer:
The refractive index of a medium depends on
1) the nature of the medium
2) the colour or wavelength of the incident light
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Question (5):
Define angle of incidence. |
| Answer:
The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. |
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Question (6):
What is the angle of incidence if a ray of light is incident normal to the surface separating the two media? |
| Answer:
Angle of incidence is equal to zero if a ray of light is incident normal to the surface separating the two media. |
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Question (7):
What is a lens? |
| Answer:
A lens is a portion of a transparent refracting medium bounded by two spherical surfaces which are generally spherical or cylindrical or one curved and one plane surface. |
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Question (8):
What is a concave lens? |
| Answer:
A lens which is thinner at the middle and thicker at the edges is called a concave lens. |
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Question (9):
What is the nature of the focus of a concave lens? |
| Answer:
The focus of a concave lens is virtual. |
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Question (10):
What type of image is formed by a concave lens? |
| Answer:
A concave lens always forms a virtual and erect image. |
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Question (11):
A thin lens has a focal length f = -12 cm. Is it convex or concave lens? |
| Answer:
The lens is concave since the focal length is negative. |
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Question (12):
A lens forms an erect image for all positions of the object in front of it. Is the lens convex or concave? |
| Answer:
Concave lens. |
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Question (13):
Where should an object be placed so that a real and inverted image of same size is obtained using a convex lens? |
| Answer:
The object has to be placed at 2F to get a real and inverted image of same size. |
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Question (14):
Write the relation between u,v and f of a thin lens.
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Answer:
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Question (15):
What is the sign of u, v and f for a convex lens according to Cartesian sign convention? |
| Answer:
According to sign convention u is negative, v is positive for all positions of the object except when the object is between the optic centre and first focus and f is positive. |
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Question (16):
An object of height 1m is placed at a distance of 2f from a convex lens. What is the height of the image formed? |
| Answer:
The height will also be equal to 1m since the object placed at 2F of a convex lens gives an image of the same at 2F on the other side of the lens. |
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Question (17):
Define power. |
| Answer:
The power of a lens is defined as the reciprocal of its focal length in metres. |
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Question (18):
What is least distance of distinct vision? |
| Answer:
The minimum distance upto which an eye can see clearly is called the least distance of distinct vision. |
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Question (19):
What happens when a ray of light passes through the optical centre of a lens? |
| Answer:
The ray of light does not suffer any deviation. |
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Question (20):
State the laws of refraction. |
Answer:
1) The incident ray, the refracted ray and the normal at the point of incidence all lie in one plane.
2) For any two given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.
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Question (21):
Diagrammatically represent the refraction of light through a rectangular glass slab. |
Answer:
 |
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Question (22):
Define convex lens. Why is it referred to as converging lens? |
| Answer:
A lens which is thicker in the middle and thinner at the edges is called convex lens.
Convex lens is referred to as a converging lens because the parallel rays of light after refraction through a convex lens meet at a point on the principal axis.
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Question (23):
Draw a diagram to show the second principal focus of a convex lens. |
Answer:
 |
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Question (24):
Distinguish between a convex and a concave lens. |
Answer:
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| Is thicker in the middle and thinner at the edges
| Is thinner at the middle and thicker at the edges
| | Focus is real
| Focus is virtual
| | It is a converging lens
| It is a diverging lens
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Question (25):
Draw a ray diagram to show the refraction of light when it passes through the optic centre of a convex lens. |
Answer:
 |
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Question (26):
List out the uses of convex lenses. |
| Answer:
Convex lenses are used
a) as a magnifying glass
b) in photocopying cameras
c) as the objective lens of a microscope and a telescope
d) in theatre spot lights
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Question (27):
With the help of a ray diagram show how an object gets magnified in a simple microscope. |
Answer:
When an object is placed between O to F1 we get an enlarged and erect image of the object.
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Question (28):
Define the power of a lens. What is its unit? |
| Answer:
Power of a lens is defined as the reciprocal of its focal length. The unit of power is dioptre. |
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Question (29):
With the help of a diagram explain how light gets refracted when it passes through a rectangular glass slab. |
Answer:
- Place a rectangular glass slab on a white sheet of paper fixed on a drawing board.
- Trace the boundary ABCD of the glass slab.
- Remove the glass slab. Draw an incident ray IO on AB.
- Draw the normal NN1 at the point of incidence O
- Fix two pins P and Q on the incident ray IO.
- Place the glass slab within its boundary ABCD.
- Looking from the other side of the glass slab fix two pins R and S such that your eye and the feet of all the pins are in one straight line.
- Remove the glass slab and the pins. Mark the pin points P1, P2, P3 and P4.
- Join OO1.It is the refracted ray.
- Measure
 are the angle of incidence, angle of refraction and angle of emergence respectively.
 - Extend O1E backwards. The emergent ray is parallel to the incident ray.
The above experiment shows that
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Answer:
The distance between the object and the lens (u) = -50 cm
Focal length f = -20 cm
Distance of the image from the optic centre = v
The image is formed 14.3 cm from the lens on the same side as the object and since v is negative the image formed is virtual and erect.
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Question (31):
An object is placed 50 cm from a lens which produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image and calculate the focal length of the lens. |
Answer:
Distance between the object and the lens = -50cm
Distance between the image and the lens = -10cm
Focal length = ?
The focal length of the lens is -12.5 cm and the negative sign indicates that the lens is concave.
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Question (32):
An object of height 4 cm is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image. |
Answer:
Distance between the object and the lens u = -10cm
Focal length = f = 20 cm
Distance between the image and the lens v = ?
The image is formed on the same side as the object and since v is negative the image is erect and virtual and is formed at a distance 20 cm in front of the lens.
We know that
height of the object ho = 4 cm
Height of the image = 8 cm |
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Question (33):
What is the power of a lens having a focal length of
a) 50 cm b) -50cm
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Answer:
When f = 50 cm
When f = -50 cm
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Question (34):
Draw a ray diagram to show the position and nature of the image formed by a convex lens when the object is placed
a) at 2F1
b) between F1 and 2F1
c) beyond 2F1
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Answer:
a) at 2F1
b) between F1 and 2F1
c) beyond 2F1
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Question (35):
State and verify Snell's law. |
Answer:
Snell's law states that for any two given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. Place a rectangular glass slab on white sheet of paper fixed on a drawing board.
Trace the boundary ABCD of the glass slab.
Remove the glass slab and draw a normal N
1
N
2
at O.
Draw a straight line IO inclined at an angle say 30
o
with the normal. IO is the incident ray.
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Fix two pins P and Q on the incident ray IO.
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Place the glass slab within its boundary ABCD.
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Looking from the other side of the glass slab fix two other pins R and S such that P, Q, R and S appear to lie on the same straight line.
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Remove the glass slab and the pins. Mark the pin points P, Q, R and S.
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Join the pins R and S and produce the line on both sides. The ray O
1
E is the emergent ray.
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Join OO
1
. It is the refracted ray.
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With O as centre, draw a circle of a convenient radius 'r' in such a way that it cuts the incident and the refracted rays at F and G respectively.
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From F and G draw perpendiculars to the normal N1N2.
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D FHO and
D GKO are right angled triangles.
- Measure the length of FH and GK.
- Repeat the experiment for different values of angle of incidence.
- Record the result in a tabular form
Find the values of
for different values of I.
Will be a constant verifying
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But when the ray of light is passing from glass to air, that is, from a denser to a rarer medium the refracted ray bends away from the normal. In this case 
