A) 17 cm to 500 cm
B) 34 cm to 250 cm
C) 20 cm to 200 cm
D) None of the above
Correct Answer: A
Solution :
From lens formula \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] ... (i) Also, power (P) \[=\frac{1}{f}\] ... (ii) From Eq. (ii), we get \[f=\frac{1}{0.8}m=-125\,\,cm\] For near point, \[{{v}_{1}}=-15\,cm,\,\,\,{{\mu }_{1}}=?\] \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] \[\frac{1}{25}=\frac{1}{-15}-\frac{1}{{{u}_{1}}}\] \[\Rightarrow \] \[\frac{1}{{{u}_{1}}}=-\frac{1}{15}+\frac{1}{125}\] \[=\frac{-125+15}{15\times 125}\] \[=-\frac{110}{125\times 15}\] \[\Rightarrow \] \[{{\mu }_{1}}=-17\,cm\] F at distant point \[{{v}_{2}}=-100\,cm,\,\,{{\mu }_{2}}=?\] \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] \[-\frac{1}{125}=\frac{1}{-100}-\frac{1}{{{u}_{2}}}\] \[\frac{1}{{{u}_{2}}}=-\frac{1}{100}+\frac{1}{125}\] \[{{u}_{2}}=-500\,\,cm\]. Hence, the person can now see objects lying between 17 cm to 500 cm.
You need to login to perform this action.
You will be redirected in
3 sec
