Question

A plane mirror is placed at a distance 12.5 cm from
focus of a concave mirror of radius of curvature 20
cm. Find where an object can be placed between
the two mirrors, so that the first image in both the
mirrors coincide.
​

Answer 1

focal length of concave mirror, f = -10cm

a plane mirror is placed at a distance of 12.5 cm from the focus of concave mirror.

Let an object is placed x distance from plane mirror.

then, image distance from plane mirror = x cm right side of plane mirror.

so, image distance , v = +x or -(12.5 + 10 - x) from concave mirror

again,

object distance from concave mirror, u = -(x + 12.5 + 10) = -(x+ 22.5)cm

using mirror formula,

1/v - 1/u = 1/f

or, 1/v - 1/-(x + 22.5) = 1/-10

or, 1/v = -1/(x + 22.5) - 1/10

or, 1/v = -(x + 32.5)/10(x + 22.5)

but v = -(22.5 - x)cm

or, -(22.5 - x) = -(x + 32.5)/10(x + 22.5)

or, 10(22.5² - x²) = (x + 32.5)

or, 5062.5- 10x² = x + 32.5

or, 10x² + x - 5030 = 0

after solving we get,

x = ± 22.378 cm

hence,object is placed 22.378 m left side from plane mirror.

Answer 2

Answer:22.378 cm

Explanation:

focal length of concave mirror, f = -10cm

a plane mirror is placed at a distance of 12.5 cm from the focus of concave mirror.

Let an object is placed x distance from plane mirror.

then, image distance from plane mirror = x cm right side of plane mirror.

so, image distance , v = +x or -(12.5 + 10 - x) from concave mirror

again,

object distance from concave mirror, u = -(x + 12.5 + 10) = -(x+ 22.5)cm

using mirror formula,

1/v - 1/u = 1/f

or, 1/v - 1/-(x + 22.5) = 1/-10

or, 1/v = -1/(x + 22.5) - 1/10

or, 1/v = -(x + 32.5)/10(x + 22.5)

but v = -(22.5 - x)cm

or, -(22.5 - x) = -(x + 32.5)/10(x + 22.5)

or, 10(22.5² - x²) = (x + 32.5)

or, 5062.5- 10x² = x + 32.5

or, 10x² + x - 5030 = 0

after solving we get,

x = ± 22.378 cm

hence,object is placed 22.378 cm left side from plane mirror.