An object of height 4.5 cm is placed 32 cm in front of a concave mirror with a radius ofcurvature of 19 cm.(i) Draw a ray diagram to show the location of the image formed by the mirror andstate whether the image is real or virtual. [3 marks](ii) Calculate the distance of the image from the mirror. [3 marks](iii) Calculate the magnification of the image

(i) As an AI, I'm unable to draw a ray diagram. However, I can explain how to draw it.

1. Draw a straight line to represent the principal axis.
2. Draw the concave mirror on the principal axis. The mirror's center of curvature (C) is 19 cm from the mirror, so mark this point on the principal axis.
3. Draw the object 32 cm in front of the mirror along the principal axis.
4. Draw a ray parallel to the principal axis from the top of the object to the mirror. This ray will reflect through the focal point (F).
5. Draw another ray from the top of the object through the center of curvature (C). This ray will reflect back along its own path.
6. The point where the two reflected rays intersect is the location of the image.

The image is real because the rays actually intersect at the image location.

(ii) To calculate the distance of the image from the mirror, we can use the mirror formula: 1/f = 1/v + 1/u.

The focal length (f) of the mirror is half the radius of curvature, so f = 19/2 = 9.5 cm. The object distance (u) is -32 cm (negative because it's in front of the mirror).

Substituting these values into the mirror formula gives us: 1/v = 1/f + 1/u = 1/9.5 - 1/(-32) = 0.105 - (-0.031) = 0.136.

So, the image distance (v) = 1/0.136 = 7.35 cm. The image is located 7.35 cm behind the mirror.

(iii) The magnification (m) of the image can be calculated using the formula m = -v/u = -7.35/(-32) = 0.23.

This means the image is 0.23 times the size of the object, or it's reduced in size by about 77%.