Question

A very thin rod of length 5 cm is placed along the axis of a concave mirror having radius of curvature 30cm. If one end of its magnified image touches an end of the rod then length of image of rod is?
a) 2.5 cm
b) 5 cm
c) 7.5 cm
4) 10 cm

Please explain with solution

Answer 1

 If an image is at the same location as the object, then the object is located at the centre of curvature.  

{ Check it : focus is at 30/2 = 15 cm.  

. . . . . . . . . . so 1/u + 1/v = 1/15  

. . . . . . . . . . and v = u, therefore 2/u = 1/15, hence u = 2*15 = 30 }  

So if one end of the rod is at 30 cm, then the other end is at 25 cm.  

And for u = 25,  

1/25 + 1/v = 1/15  

1/v = 1/15 - 1/25  

v = 75/2 = 37.5 cm  

So one end of the image is located at 30 cm, and the other end is located at 37.5 cm.