zigya tab
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Here,                    P = + 1.5 D
therefore                         straight f space equals space 1 over straight P space equals space fraction numerator 1 over denominator plus 1.5 space straight D end fraction space equals space plus 10 over 15 space straight m space equals space plus space 0.67 space cm space space space Ans. space
As the focal length is positive, the prescribed lens is converging.
Spherical mirrors are of two types:
(i) Concave mirror: A spherical mirror is concave if it is silvered on the outer bulged surface and reflects light from the inner hollow surface [Fig. (b)]

Here,                    P = + 1.5 D               .
Fig. (a) A spherical shell cut by a plane surface, (b) concave mirror, and (c) convex mirror.
(ii) Convex mirror: A spherical mirror is convex if it is silvered on the inner hollow surface and reflects light from the outer bulged surface.



An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and the nature of the image.

We are given a concave mirror. 

Object size, h = + 7.0 cm
Object distance, u = - 27 cm
Focal length, f = - 18 cm 
Image distance, v = ?

Image size, h' = ? 

Now, using the mirror formula, 

                     1u+1v = 1f

                  1v = 1f-1u 

                      = 1-18-1-27 = -3+254 = -154

i.e.,                  v = -54 cm 

The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image. 

Magnification, m = h'h = -vu 

 Image size, 
                        h' = -vhu   

                           =-(-54)×(+7)(-27)= -14 cm 

The image is real, inverted and enlarged in size.