A point source of light is placed at a distance, of 2 f from a converging lens of focal length 1 The intensity on the othe

The power of a lens is related to its focal length as P = \(\frac{1}{f}\)(in m) More the power of a lens more is its converging power. Thus, the lens of shorter focal...

Given P=+ 1.5D As P=1/f (in metre) , Focal length, f =1/P=1/1.5=0.667m=66.7cm As power of prescribed lens is positive, the lens is converging (or convex) lens.

The image is real, inverted and diminished as the object is placed beyond 2F.

P=1/f , P∝1/f. Power of a lens is inversely proportional to its focal length therefore lens having focal length of 20 cm will provide more convergence.

(a) point source EXPLANATION:  Only for a point source the intensity can be written as I = E/4πr², where E is the strength of the source. Hence  E ∝ 1/r².

(a) must be less than 10 cm

The correct answer is (a) =x

The correct answer is (c) fv < fr

(b) the image will be shifted a little later than the object

Let focal length of converging and diverging lenses be + f and - f respectively. Power of converging len P1 = 1/f, Power of diverging lens P2 = - 1/f ∴ Power of combination P...