Question

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 1.0 cm high what will be the size of image





( can you make the figure plz)​

Answer 1

Answer:

focul length is 10 cm, f= -10( focul lenght of concave mirror is negative)

so, radius in 20 cm

object distance u is 50 cm, so the object is beyond C( centre of curvature)

hence u= -50( distance on the left of pole, and vertically below the principal axis are negative)

mirror formula is

1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = -(1/10) - {-(1/50)}

1/v = -(1/10) + (1/50)

1/v = 1/50 - 1/10

1/v = - 4/50

so, v = - 50/4 = -12.5 cm

So, image is real and inverted, and between C & F

now, magnification, m = h’/h = -v/u

m = -v/u = -(-12.5/-50) = -1/4

So, h’/h = - 1/4

h, height of object = 1 cm

so, h’/1 = - 1/4

h’, height of image = -0.25 cm

height is negative as the image is real and inverted.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Therefore, height of the image is -0.166 cm, which means that the image is inverted, highly diminished and is a real image. therefore the height of the image formed is 0.25cm... The concave mirror focal length is 18 and the image of an object kept in the mirror is half the height of objects.