Question

An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and $\mu =1.5$ is then placed close to the mirror in the space between the object and the mirror. The position of final image formed is

• A. $-3.94cm$
• B. $4.3cm$
• C. $-4.93cm$
• D. $3.94cm$

Answer 1

The correct option is C $-4.93cm$

Due to glass slab, increase in path

$=(\mu -1)t$

$\left\{\begin{array}{c}\mu =re\text{fractive index}\\ t=\text{thickness of slab}\end{array}\right\}$

$\begin{array}{cc}\text{torease path}& =(1.5-1)3\\ t& =1.5cm\end{array}$

new initial distance =

$\begin{array}{cc}{u}^{\prime }& =-(21+1.5)\\ & =-22.5cm\end{array}$

$\begin{array}{cc}\text{Now,}\frac{1}{-22.5}+\frac{1}{v}=\frac{1}{-5}& \\ \Rightarrow \frac{1}{v}& =\frac{1}{22.5}-\frac{1}{5}\\ \Rightarrow \frac{1}{v}& =\frac{5-22.5}{22.5\times 5}\\ \therefore & v=\frac{22.5\times 5}{-17.5}=-6.43cm\\ \Rightarrow & \vee =-6.43cm\end{array}$

$\begin{array}{cc}\text{Geometrical distance}& =-(6.43-1.5)\\ & =-4.93cm\end{array}$