1. A small pin is placed 10 cm in front of a concave mirror of focal length 15 cm. calculate the image distance and the magnification of the image .what is the nature of the image formed.
2. A concave mirror of focal length 15 cm forms an image of a pin 12 cm high at 20 cm in front of the mirror. find the object distance& magnification. Find the size of the image.
3. The height of an object placed at a distance of 50 cm from the hole of a pinhole camera is 10 cm. if the length of the camera is 8 cm. calculate the magnification produced by the camera and the height of the image.
4. A converging lens of focal length 15 cm is used to obtain a real image magnified one and a half times. calculate the distance of the image from the lens..
5. An object is placed 30 cm from a diverging lens of focal length 15 cm calculate the distance of the image from the lens.
6. An object is placed 30 cm in front of a concave mirror of focal length 20 cm. calculate the distance between the image and the objects.
7. A man 1.8 m tall stands 3 m away from a pinhole camera. if the distance between the pinhole and the screen of the camera is 0.3 m. calculate the height of the image of the man produced by the camera.
8. An object placed at cm from a lens has its real image located 48 cm from the lens .calculate the focal length of the lens.
9. The real image of an object formed by a converging lens of focal length 15 cm is three times the size of the object .
calculate the object distance.
10. A magnified virtual image four times the size of the object is formed by a concave mirror of focal length 12 cm what is the distance of the image from the pole of the mirror.
1. Using the mirror formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.
Given f = 15 cm and u = -10 cm (since the object is placed in front of the mirror),
1/15 = 1/v - 1/-10
1/15 = (10-v) / 10v
10v = 15(10 - v)
10v = 150 - 15v
25v = 150
v = 6 cm
The image distance is 6 cm.
To calculate the magnification, we use the formula m = -v/u.
m = -6 / -10
m = 0.6
The magnification is 0.6.
Since the image is formed in front of the mirror (i.e., v is positive), the nature of the image is virtual and erect.
2. Using the mirror formula again, 1/f = 1/v - 1/u.
Given f = 15 cm and v = 20 cm,
1/15 = 1/20 - 1/u
1/15 = (u - 20) / 20u
20u = 15(u - 20)
20u = 15u - 300
5u = 300
u = 60 cm
The object distance is 60 cm.
To calculate the magnification, we use the formula m = -v/u.
m = -20 / 60
m = -1/3
The magnification is -1/3.
To find the size of the image, we multiply the height of the object by the magnification.
Size of the image = (12 cm) * (-1/3)
Size of the image = -4 cm
The size of the image is -4 cm.
3. The magnification produced by the pinhole camera is given by the formula m = h'/h, where h' is the height of the image and h is the height of the object.
Given h = 10 cm,
m = h'/10
We need to find h', the height of the image.
Using similar triangles, we can set up the equation h'/h = (l - d)/d, where l is the length of the camera and d is the distance between the hole and the image (which is the same as the distance between the hole and the screen).
Given l = 8 cm and d = 0.3 m,
h'/10 = (8 - 0.3) / 0.3
h'/10 = 7.7 / 0.3
h' = (7.7 / 0.3) * 10
h' = 256.67 cm
The height of the image is approximately 256.67 cm.
4. Using the lens formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.
Given f = 15 cm and m = 1.5 (since the image is magnified one and a half times),
m = -v/u
1.5 = -v/u
v = -1.5u
Substituting this into the lens formula, we get:
1/15 = 1/(-1.5u) - 1/u
1/15 = (-1 + 1.5)/(-1.5u)
1/15 = 0.5/(-1.5u)
0.5u = -1.5(15)
0.5u = -22.5
u = -45 cm
The object distance is -45 cm.
To calculate the distance of the image from the lens, we use the equation v = -1.5u.
v = -1.5(-45)
v = 67.5 cm
The distance of the image from the lens is 67.5 cm.
5. Using the lens formula again, 1/f = 1/v - 1/u,
Given f = -15 cm (since it's a diverging lens) and u = -30 cm,
1/-15 = 1/v - 1/-30
-2/30 = (30 + v) / 30v
-2/30 = (1 + v/30) / v
-1/15 = (1 + v/30) / v
-1/15 = (v + 30) / 30v
30v = -15(v + 30)
30v = -15v - 450
45v = -450
v = -10 cm
The distance of the image from the lens is -10 cm.
6. Using the mirror formula, 1/f = 1/v - 1/u,
Given f = -20 cm (since it's a concave mirror) and u = -30 cm,
1/-20 = 1/v - 1/-30
-3/60 = (30 - v) / -30v
-1/20 = (30 - v) / -v
1/20 = (v - 30) / v
1/20 = (v - 30) / (v - 30)
1 = v - 30
v = 31 cm
The distance between the image and the object is 31 cm.
7. The height of the image produced by the pinhole camera is given by the formula h' = h * (d'/d), where h' is the height of the image, h is the height of the object, d' is the distance between the pinhole and the screen (which is the same as the distance between the pinhole and the image), and d is the distance between the pinhole and the object.
Given h = 1.8 m, d = 3 m, and d' = 0.3 m,
h' = 1.8 * (0.3/3)
h' = 0.6 m
The height of the image produced by the camera is 0.6 m.
8. Using the lens formula, 1/f = 1/v - 1/u,
Given v = -48 cm and u = -x cm,
1/f = 1/v - 1/u
1/x = 1/-48 - 1/u
1/x + 1/u = 1/-48
(1 + x/u)/(xu) = -1/48
(x + u)/(xu) = -1/48
48(x + u) = -xu
48x + 48u = -xu
48x = -xu - 48u
u(x + 48) = -48x
u = -48x / (x + 48)
The focal length of the lens is given by -48x / (x + 48).
To find x, we can set up the equation: -48(x + 48) = 48x.
Solving for x, we get x = -24.
Substituting this back into the equation for u, we get:
u = -48(-24) / (-24 + 48)
u = 48 cm
The object distance is 48 cm.
9. The magnification produced by the converging lens is given by the formula m = -v/u, where m is the magnification, v is the image distance, and u is the object distance.
Given f = 15 cm and m = 3 (since the image is three times the size of the object),
m = -v/u
3 = -v/u
v = -3u
Substituting this into the lens formula, we get:
1/15 = 1/(-3u) - 1/u
1/15 = (-3 + 1)/(-3u)
1/15 = -2/(-3u)
3u = -2(15)
3u = -30
u = -10 cm
The object distance is -10 cm.
10. Using the mirror formula for a concave mirror, 1/f = 1/v - 1/u,
Given f = -12 cm (since it's a concave mirror) and m = 4 (since the image is four times the size of the object),
m = -v/u
4 = -v/u
v = -4u
Substituting this into the mirror formula, we get:
1/-12 = 1/(-4u) - 1/u
-1/12 = (-4 + 1)/(-4u)
-1/12 = -3/(-4u)
4u = -3(12)
4u = -36
u = -9 cm
The object distance is -9 cm.