question_answer

Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

The distance between two point object 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? from the lens could be (1) 20 cm                              (2) 12 cm (3) 18 cm                              (4) 16 cm

A)  1, 2 and 3 are correct

B)  1 and 2 are correct

C)   2 and 4 are correct

D)  1 and 3 are correct

Correct Answer: B

Solution :

                 Since images are formed at the same place, so one image must be real and other must be virtual.                                 \[x+y=32\]                 For P                 \[\frac{1}{y}-\frac{1}{-x}=\frac{1}{15}\]                 Or           \[\frac{1}{x}+\frac{1}{y}=\frac{1}{15}\]                  ?.(i)                 For Q                 \[\frac{1}{-y}-\frac{1}{-(32-x)}=\frac{1}{15}\]                 \[-\frac{1}{y}+\frac{1}{(32-x)}=\frac{1}{15}\]       ?..(ii)                 Adding Eqs. (i) and (ii), we get                 \[\frac{1}{x}+\frac{1}{(32-x)}=\frac{2}{15}\]                 \[\frac{32}{x(32-x)}=\frac{2}{15}\]                 or         \[32x-{{x}^{2}}=240\]                 or            \[{{x}^{2}}-32x+240=0\]                 or            \[(x-20)(x-12)=0\]                 \[x=12cm\]and\[x=20\text{ }cm\]