Home/Class 12/Physics/

## QuestionPhysicsClass 12

Find the position of the image formed by the lens combination given in the figure.

For $$L_1$$ $$($$lens on left$$)$$
$$u_1=-30\;\text{cm}$$ , $$f_1=10\text{ cm}$$
From lens formula,
$$\frac 1{f_1}=\frac 1{v_1}-\frac 1{u_1}$$
$$\Rightarrow v_1=15\text{ cm}$$
This will act as virtual object for $$L_2$$ .
Now for $$L_2$$ ,
$$u_2=10\text{ cm}$$    $$\{$$distance of image by $$L_1$$ from $$L_2\}$$
$$f_2=-10\text{ cm}$$
By using lens formula we get,
$$v_2={\infty}$$
This will act as virtual object for $$L_3$$ .
Now for $$L_3$$ ,
$$f_3=30\text{ cm}$$ , $$u_3={\infty}$$
By using lens formula we get,
$$v_3=30\text{ cm}$$
$${\therefore}$$  The position of the image by combination of lens is $$45\text{ cm}$$  from $$L_1$$ $$[$$on the right$$]$$ or $$30\text{ cm}$$  from $$L_3$$  $$[$$on the right$$]$$