A) convergent lens of focal length 3.5R
B) convergent lens of focal length 3.07R
C) divergent lens of focal length 3,5R
D) divergent lens of focal length 3.07R
Correct Answer: A
Solution :
[a] Use lens maker's formula \[\frac{1}{f}=\left( _{g}^{m}\mu -1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] Now, \[_{g}^{m}\mu =\frac{_{g}\mu }{_{m}\mu }=\frac{1.5}{1.75}\] \[\therefore \frac{1}{f}=\left( \frac{1.5}{1.75}-1 \right)\left( -\frac{1}{R}-\frac{1}{R} \right)=+\frac{0.25\times 2}{1.75R}\] \[\Rightarrow f+3.5R\] The positive sign shows that the lens behaves as convergent lens.
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