a concave mirror of focal length 10 cm is kept in front of an object at a distance of 50 cm in front of it if the object 1 cm high what will be the size of image
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f = -10 cm
u = -50 cm
h1 = 1 cm
v = ?
so , 1/f = 1/v + 1/u
1/ -10 = 1/v + 1/ -50
-1/10 = 1/v - 1/50
-1/10 + 1/50 = 1/v
-5+1/50 = 1/v
-4/50 = 1/v
-2/25 = 1/v
so v= - 25/2 cm
now h1/h2 = -v/u
putting the values
1/h2 = - ( -25/2) / 50
1/h2 = 25/2 * 1/50
1/h2 = 1/4
therefore h2 = 4 cm
u = -50 cm
h1 = 1 cm
v = ?
so , 1/f = 1/v + 1/u
1/ -10 = 1/v + 1/ -50
-1/10 = 1/v - 1/50
-1/10 + 1/50 = 1/v
-5+1/50 = 1/v
-4/50 = 1/v
-2/25 = 1/v
so v= - 25/2 cm
now h1/h2 = -v/u
putting the values
1/h2 = - ( -25/2) / 50
1/h2 = 25/2 * 1/50
1/h2 = 1/4
therefore h2 = 4 cm
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