IF SOMEONE IS ONLINE PLEASE ANS THESE Qs FASSTT!! GOT BOARDSSS!!!!!
1. Two thin lenses of focal lengths 10 cm and -5 cm are kept in contact. What is the focal length and power of the combination?
2. An image 2 cm high is placed at a distance of 16 cm from a concave mirror which produces a real image 3 cm high. Find the position of the image and the focal length of the mirror.
3. An object placed at a distance of 10 cm from a concave mirror produced an infinitely large image at the infinity. The focal length of the mirror is?
4. A concave mirror produces a real inverted image of the same size as that of the object. If the object is placed at a distance of 16 cm from the mirror, the focal length of the mirror is?
Answers
Answered by
6
1. f1 = 10 cm = 0.1 m f2 = - 5 cm = - 0.05 m
P1 = power of first lens = 1/f1 = 10 Dioptres
P2 = 1/f2 = - 20 dioptres
combination power = P = P1 + P2 = 10 - 20 Dioptres = -10 D
combination focal length = 1/ P = -1/10 m = -10 cm
2. h = 2cm u = -16 cm
h' = 3 cm erect or - 3 cm inverted
case 1:
m = h'/h = 1.5 = -v/u
v = - 1.5 u = 24 cm
1/f = 1/v+ 1/u = 1/24 - 1/16 = -1/48
f = - 48 cm
case 2 :
h' = -3 cm m = -1.5
v = 1.5 u = - 24 cm
1/f = -1/24 + 1/16 = 1/48
f = 48 cm
3. u = -10 cm v = infinity m = infinity
1/f = 1/u + 1/v = -1/10 + 1/infinity = -1/10
f = -10 cm
When an object is placed at the focus of the mirror, it produces an image at infinity. When parallel rays from infinite distance fall on the mirror, they produce an image at the focus.
4. h' = - h m = -1
u = -16 cm => v = - m u = - 16 cm
1/f = -1/16 -1/16 = -1/8
f = - 8 cm
When an object is placed at the center of curvature C of the concave mirror with focal length f, then it produces an image inverted at the same place ie., center of curvature.
then u = - R = 2 f hence, f = - R/2 = u/2
P1 = power of first lens = 1/f1 = 10 Dioptres
P2 = 1/f2 = - 20 dioptres
combination power = P = P1 + P2 = 10 - 20 Dioptres = -10 D
combination focal length = 1/ P = -1/10 m = -10 cm
2. h = 2cm u = -16 cm
h' = 3 cm erect or - 3 cm inverted
case 1:
m = h'/h = 1.5 = -v/u
v = - 1.5 u = 24 cm
1/f = 1/v+ 1/u = 1/24 - 1/16 = -1/48
f = - 48 cm
case 2 :
h' = -3 cm m = -1.5
v = 1.5 u = - 24 cm
1/f = -1/24 + 1/16 = 1/48
f = 48 cm
3. u = -10 cm v = infinity m = infinity
1/f = 1/u + 1/v = -1/10 + 1/infinity = -1/10
f = -10 cm
When an object is placed at the focus of the mirror, it produces an image at infinity. When parallel rays from infinite distance fall on the mirror, they produce an image at the focus.
4. h' = - h m = -1
u = -16 cm => v = - m u = - 16 cm
1/f = -1/16 -1/16 = -1/8
f = - 8 cm
When an object is placed at the center of curvature C of the concave mirror with focal length f, then it produces an image inverted at the same place ie., center of curvature.
then u = - R = 2 f hence, f = - R/2 = u/2
Similar questions