Answer:
Given: \[{{h}_{1}}=+5\text{ }cm,f=-10\text{ }cm,u=-\text{ }20\text{ }cm\] we know that, \[\frac{1}{f}=\frac{1}{\upsilon }-\frac{1}{u}\] Or \[\frac{1}{\upsilon }=\frac{1}{f}+\frac{1}{u}=\frac{1}{-10}-\frac{1}{20}=\frac{-3}{20}\] Image distance, \[\upsilon =-\frac{20}{3}cm\] The nature of the image is virtual and erect. Now, magnification, \[m=\frac{{{h}_{2}}}{{{h}_{1}}}=\frac{\upsilon }{u}\] \[\Rightarrow \] \[{{h}_{2}}=\frac{\upsilon }{u}\times {{h}_{1}}=\frac{-20}{3}\times \frac{1}{-20}\times 5=\frac{+5}{3}cm\] \[\therefore \] The size of the image is 1.67 cm.
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