Question

The focal length of a given lens is -51 cm. Given that the image is virtual, erect, and 10 percent of the size of the object.

Now complete the following statements: 

The position of the object is ?

The position of the image is ?

Answer 1

focal length $(f)=-51 cm$

Magnification(m)$=\dfrac{y_i}{y_o}=\dfrac{10}{100}=\dfrac{1}{10}$

$\dfrac{v}{u}=\dfrac{1}{10}~~~~-(1)$

Using the lens formula-

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\[9pt] \Rightarrow \dfrac{1}{-51}=\dfrac{10}{u}-\dfrac{1}{u} \\[9pt] \Rightarrow \dfrac{-1}{51}=\dfrac{9}{u} \\[9pt] \Rightarrow u=-\dfrac{51}{9}=-5.66 cm$

So, $v=\dfrac{u}{10}=\dfrac{-5.66}{10}=-0.566 cm$

Hence Position of the object is at -5.66cm and position of image is at -0.566cm.