The size of image will be
A) -0.4 cm
B) -0.5 cm
C) 0.6 cm
D) 0.7 cm
E) 3.0 cm
Correct Answer: C
Solution :
From the figure, \[u=-\,40\,cm,\,R=-\,20\,cm\] \[{{\mu }_{1}}=1,\]\[{{\mu }_{2}}=133,\] we have, \[\frac{{{\mu }_{2}}}{v}-\frac{{{\mu }_{4}}}{u}=\frac{{{\mu }_{2}}-{{\mu }_{4}}}{R}\] \[\Rightarrow \] \[\frac{133}{v}-\frac{1}{-\,40\,cm}=\frac{133-1}{-\,20\,cm}\] \[\Rightarrow \] \[v=-\,32\,cm\] The magnification is \[m=\frac{{{h}_{2}}}{{{h}_{1}}}=\frac{{{\mu }_{1}}v}{{{\mu }_{2}}u}\] \[\Rightarrow \] \[\frac{{{h}_{2}}}{1.0\,cm}=\frac{-\,32\,cm}{1.33\times (-\,40\,cm)}\] \[\Rightarrow \] \[{{h}_{2}}=+\,0.6\,cm\] Positive sign shows that the image is erect.
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