A small insect is placed in front of a concave mirror (radius = 5.50cm), such th

Question

A small insect is placed in front of a concave mirror (radius = 5.50cm), such that the image distance equals the object distance. How far is the insect from the mirror? A dentist wants a small mirror that, when 2.12cm from a tooth, will produce a 4.09x upright image, what is the mirrors radius of curvature? Answer: An object whose height is 5.1 cm is at a distance of 12.5 cm from a spherical concave mirror. Its image is real and has a height of 10.1 cm. Calculate the radius of curvature of the mirror. Correct, computer geta, 1.66 z + 01 cm How far from the mirror is it necessary to place the above object in order to have a virtual image with a height of 10.1 cm? Answer: Last Answer: 12.40 cm Incorrect, tries 3/6.

Explanation / Answer

4)

when object distance is r, image distance will be r.

so, object distance from the mirror = r

= 5.50 cm

5)

object distance, u = 2.12 cm

m = 4.09

let v is the image distance.

we know, m = -v/u

==> v = -m*u

= -4.09*2.12

= -8.67 cm

now use, 1/f = 1/u + 1/v

1/f = 1/2.12 + 1/(-8.67)

f = 2.81 cm

r = 2*f

= 2*2.81

= 5.62 cm

6)

object distance, u = 12.5 cm

m = -10.1/5.1

= -1.98

let v is the image distance.

we know, m = -v/u

v = -m*u

= -(-1.98)*12.5

= 24.75 cm

now use, 1/f = 1/u + 1/v

1/f = 1/12.5 + 1/24.75

f = 8.31 cm

r = 2*f

= 2*8.31

= 16.6 cm

7) for virtual image, m = +1.98

m = -v/u

v = -m*u

= -1.98*u

now use,

1/u + 1/v = 1/f

1/u + 1/(-1.98*u) = 1/16.6

==> u = 8.22 cm

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