question_answer

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 4 m away by means of a large convex lens. The maximum possible focal length of the lens required for this purpose will be:

A)  0.5m                                    

B)  1.0m

C)   1.5 m                                  

D)  2.0 m

Correct Answer: B

Solution :

                 Let a large convex lens is placed between two walls at a distance x from wall on which an electric bulb is fixed. Using lens formula, \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] Putting\[u=x\]     and \[v=4-x\] \[\therefore \]  \[\frac{1}{f}=\frac{1}{4-x}-\frac{1}{-x}\] Or           \[\frac{1}{f}=\frac{x+4-x}{(4-x)(x)}\] Or           \[\frac{1}{f}=\frac{4}{(4-x)(x)}\] Or           \[f=\frac{(4-x)(x)}{4}\]                  ?.. (i) Now magnification, \[m=\frac{v}{u}=\frac{4-x}{x}\] or           \[1=\frac{4-x}{x}\] or            \[x=4-x\] or            \[2x=4\] or            \[x=2m\] Hence, from equation (i),                 \[f=\frac{(4-2)(2)}{4}=\frac{2\times 2}{4}=1m\]