A square ABCD of side 1mm is kept at distance 15cm infront of the concave mirror as show in the figure. The focal length of the mirror is 10 cm. The length of the perimeter of its image will be(nearly):

The problem seems to be incomplete as there is no figure provided. However, assuming that the square is placed perpendicular to the principal axis of the mirror, we can solve the problem using the mirror formula:

1/v + 1/u = 1/f

where:

• v is the image distance,
• u is the object distance, and
• f is the focal length of the mirror.

Given that u = -15 cm (the negative sign indicates that the object is on the same side of the mirror as the light source) and f = -10 cm (the negative sign indicates that the mirror is concave), we can substitute these values into the mirror formula to find v:

1/v = 1/f - 1/u 1/v = 1/(-10 cm) - 1/(-15 cm) 1/v = -0.1 cm⁻¹ + 0.067 cm⁻¹ 1/v = -0.033 cm⁻¹ v = -30 cm

The negative sign indicates that the image is on the same side of the mirror as the object.

The magnification m of the mirror is given by the ratio of the image distance to the object distance:

m = -v/u m = -(-30 cm)/(-15 cm) m = -2

The negative sign indicates that the image is inverted, and the magnitude of m indicates that the image is twice the size of the object.

Therefore, if the side of the square is 1 mm, the side of the image will be 2 mm. Since a square has four sides, the perimeter of the image will be 4 * 2 mm = 8 mm.