A) \[30\text{ }cm,\text{ }-60\text{ }cm\]
B) \[20\text{ }cm,\text{ }-30\text{ }cm\]
C) \[15\text{ }cm,\text{ }-20\text{ }cm\]
D) \[12\text{ }cm,\text{ }-15\text{ }cm\]
Correct Answer: B
Solution :
[b]| Initially, \[F=60\text{ }cm\] (Focal length of combination) | |
| Hence, by using \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] | |
| \[\Rightarrow \] \[\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}=\frac{1}{60}\Rightarrow \,\,\frac{{{f}_{1}}{{f}_{2}}}{{{f}_{1}}+{{f}_{2}}}\] | ...(i) |
| Finally by using \[\frac{1}{F'}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}-\frac{d}{{{f}_{1}}{{f}_{2}}}\] | |
| where \[F'=30\,cm\] and \[d=10\text{ }cm\] | |
| \[\Rightarrow \] \[\frac{1}{30}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}-\frac{10}{{{f}_{1}}{{f}_{2}}}\] | ...(ii) |
| From (i) and (ii), \[{{f}_{1}}{{f}_{2}}=-600\] | |
| From equation (i) \[{{f}_{1}}+{{f}_{2}}=-10\] | ...(iii) |
| Also, difference of focal lengths can written as | |
| \[{{f}_{1}}-{{f}_{2}}=\sqrt{{{({{f}_{1}}+{{f}_{2}})}^{2}}-4{{f}_{1}}{{f}_{2}}}\] | |
| \[\Rightarrow \] \[{{f}_{1}}-{{f}_{2}}=50\] | ...(iv) |
| From (iii) \[\times \] (iv), \[{{f}_{1}}=20\] and \[{{f}_{2}}=-30\] | |
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