For a concave mirror with the focal length of 4.6 cm, where should a 5.0 cm tall object be placed with a respect to a mirror so that it produces a 8.0 cm tall virtual image located 12 cm from the...

For a concave mirror with the focal length of 4.6 cm, where should a 5.0 cm tall object be placed with a respect to a mirror so that it produces a 8.0 cm tall virtual image located 12 cm from the mirror?

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llltkl | College Teacher | (Level 3) Valedictorian

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We know, a virtual and enlarged image is formed in a concave mirror when the object is placed between the mirror and its focal point. The exact location can be calculated from the mirror equation as follows:

Here, f = -4.6 cm (concave mirror).

d_i = 12 cm (image on the right side of the mirror).

Applying mirror equation for the concave mirror,

1/f = 1/d_o + 1/d_i

Putting the values, we get

-1/4.6 = 1/d_o + 1/12 ---- (i)

Therefore,

1/d_o = -1/4.6 – 1/12

= - (1/4.6 + 1/12)

= - (12+4.6)/12*4.6

Or, d_o = -12*4.6/(16.6) cm

= - 3.3 cm (the - sign indicating the distance is on the left side of the mirror).

Therefore, the object was placed at a distance of 3.3 cm, in the left side (front) of the mirror.

Sources:

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