question_answer

The distance between two point objects P and Q is 32 cm. A convex lens of focal length 15 cm is placed between them, so that the images of both the objects are formed at the same place. The distance of\[P\]from the lens could be

A)  20 cm                                  

B)  12 cm

C)  18 cm                                  

D)  28 cm

Correct Answer: A

Solution :

                \[x+y=32\] For P      \[\frac{1}{y}-\frac{1}{-x}=\frac{1}{15}\]                 \[\frac{1}{x}+\frac{1}{y}=\frac{1}{15}\] For Q     \[\frac{1}{-y}-\frac{1}{-(32-x)}=\frac{1}{15}\]                 \[-\frac{1}{y}+\frac{1}{(32-x)}=\frac{1}{15}\]       Adding Eqs.. (i) and (ii) we get \[\frac{1}{x}+\frac{1}{(32-x)}=\frac{2}{15}\]                 \[\frac{32}{x(32-x)}=\frac{2}{15}\] \[32-{{x}^{2}}=240\] \[{{x}^{2}}-32{{x}^{4}}-240=0\] \[(x-20)(x-12)=0\] \[x=12cm\] and \[x=20cm\]