A) 6 cm
B) 9 cm
C) 12 cm
D) 15 cm
Correct Answer: A
Solution :
[a] The given condition will be satisfied only if one source \[({{S}_{1}})\] placed on one side such that \[u<f\] (i.e., it lies under the focus). The other source (5y 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 | |
\[\frac{1}{f}=\frac{1}{+y}\,-\frac{1}{-(24-x)}\] | |
\[\Rightarrow \] \[\frac{1}{y}=\frac{1}{y}-\frac{1}{(24-x)}\] | ...(ii) |
From (i) and (ii), |
\[\frac{1}{x}-\frac{1}{f}=\frac{1}{f}-\frac{1}{(24-x)}\] |
\[\Rightarrow \] \[\frac{1}{x}+\frac{1}{(24-x)}=\frac{2}{f}=\frac{2}{9}\] |
\[\Rightarrow \] \[{{x}^{2}}-24x+108=0\] |
On solving the equation, \[x=18\text{ }cm,\text{ }6\text{ }cm\] |
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