(i)A concave mirror has a focal length of 20 cm.find the position of the object for which the image size is double that of object size.
(ii)A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm.Find the location size and nature of the image.
(iii)Find the diameter of the image of the moon formed by the spherical concave mirror of focal length of the moon formed by a spherical concave mirror of focal length 7.6 m.The diameter of the moon is 3450 km and distance between the earth and the moon is 3.8 * 10^5 km.
please explain all questions and say me fast
kvnmurty:
why are you writing so many questions in one question ??? you should write in three separate questions... i will delete the questions next time..
you are right, so I didnot give response of all question
Answers
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1) Image can be virtual => m = 2, v = - 2 u
1/-2u + 1/u = 1/-20 => u = -10cm
image real, m = -2, v = 2 u
1/2u + 1/u = 1/-20 => u = -30 cm
two positions
2) 1/v + 1/u = 1/f
1/v + 1/-7.5 = 1/6 => v = 3.33 cm a virtual erect image is formed.
magnification = 3.33/7.5 = 4/9 so image size = 3 * 4/9 = 4/3 cm
3)
u is so high that 1/u is nearly negligible as compared to 1/f. Hence 1/v = 1/-7.6
v = nearly - 7.6 cm.
m = - 7.6/380,000,000
- h'/h = m => h' = - 7.6 * 3450 /380,000 = - 0.069 meters
h' = 6.9 cm => diameter of the image formed by mirror at 7.6 metes or focus
1/-2u + 1/u = 1/-20 => u = -10cm
image real, m = -2, v = 2 u
1/2u + 1/u = 1/-20 => u = -30 cm
two positions
2) 1/v + 1/u = 1/f
1/v + 1/-7.5 = 1/6 => v = 3.33 cm a virtual erect image is formed.
magnification = 3.33/7.5 = 4/9 so image size = 3 * 4/9 = 4/3 cm
3)
u is so high that 1/u is nearly negligible as compared to 1/f. Hence 1/v = 1/-7.6
v = nearly - 7.6 cm.
m = - 7.6/380,000,000
- h'/h = m => h' = - 7.6 * 3450 /380,000 = - 0.069 meters
h' = 6.9 cm => diameter of the image formed by mirror at 7.6 metes or focus
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1)answer is 10cm I think
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