A) 6 cm
B) 9 cm
C) 12 cm
D) 15 cm
Correct Answer: A
Solution :
The given condition will be satisfied only if one source (S1) placed on one side such that u < f (i.e. it lies under the focus). The other source (S2) is placed on the other side of the lens such that u > f (i.e. it lies beyond the focus). If \[{{S}_{1}}\] is the object for lens then \[\frac{1}{f}=\frac{1}{-y}-\frac{1}{-x}\] \[\Rightarrow \]\[\frac{1}{y}=\frac{1}{x}-\frac{1}{f}\] .....(i) If \[{{S}_{2}}\]is the object for lens then \[\frac{1}{f}=\frac{1}{+y}-\frac{1}{-(24-x)}\]\[\Rightarrow \]\[\frac{1}{y}=\frac{1}{f}-\frac{1}{(24-x)}\] .....(ii) From equation (i) and (ii) \[\frac{1}{x}-\frac{1}{f}=\frac{1}{f}-\frac{1}{(24-x)}\]\[\Rightarrow \]\[\frac{1}{x}+\frac{1}{(24-x)}=\frac{2}{f}=\frac{2}{9}\] \[\Rightarrow \]\[{{x}^{2}}-24x+108=0\]. After solving the equation \[x=18\ cm\], 6 cm.You need to login to perform this action.
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