physics

Question

 A 5.8 cm tall object is placed 38 cm in front of a concave mirror with a radius of curvature of

62 cm.

a) What is the image distance?

b) What is the size of image produced?

c) State whether the image is: real or virtual, erect or inverted, and larger or smaller than the

object.

Answer 1

a)

Object height h_object = 5.8 cm

Object distance d_object = 38 cm

Image distance d_image = ?

Radius of curvature C = 62 cm

Focal length f= C/2 = 62/2 = 31 cm

Formula : (1/f) = (1/d_object) + (1/d_image)

(1/31) = (1/38) + (1/d_image)

(1/d_image) = (1/31) - (1/38)

(1/d_image) = (38-31)/(31*38)

(1/d_image) = (7)/(31*38)

(1/d_image) = (1/168.28)

d_image = 168.28  cm.

Solution : Image distance is 168.28 cm.

Answer 2

b)

The magnification M = - (d_image)/(d_object) = (h_image)/(h_object)

- (168.28)/(38) = (h_image)/(5.8)

h_image= - (168.28*5.8)/(38)

h_image = - 25.68 cm.

The negative values for image height indicate that the image is an inverted image.

Solution : Image size (height) is 25.68 cm.

Answer 3

c)

A concave mirror will can produce

      a) A virtual & upright image occurs if the object is located between focal point and mirror.

          Image is larger than object.

      b) A real image (same size) occurs if the object is located at Centre of curvature.

      c) A real & inverted image occurs if the object is located between focus and

         Centre of curvature. Image is larger than the object.

      d) No Image will be formed if the object is located at focal point.

      e)  A real & inverted image occurs if the Object is located beyond Centre of curvature.

          Image is smaller than the object.

Object distance is 38 cm.

Focal length is 31 cm.

Radius of curvature or Center of curvature is 62 cm.

So the object is located between focus and Centre of curvature (case(c)).

Solution : Image is real , inverted and image size is larger than the object.