Question 16: Ray Optics

Question

An object $A B$ is placed 15 cm on the left of a convex lens $P$ of focal length 10 cm . Another convex lens $Q$ is now placed 15 cm right of lens $P$. If the focal length of lens $Q$ is 15 cm , the final image is $\_\_\_\_$ .

Accepted Answer

For Lens P:
Focal length is $\mathrm{f}_{\mathrm{P}}=+10 \mathrm{~cm}$
Object distance is $\mathrm{u}_{\mathrm{P}}=-15 \mathrm{~cm}$
Using the lens formula.
$$ \frac{1}{v_P}-\frac{1}{u_P}=\frac{1}{f_P} $$
$$\Rightarrow $$ $\frac{1}{v_P}-\frac{1}{-15}=\frac{1}{10}$
$$\Rightarrow $$ $\frac{1}{v_P}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}$
$$ \mathrm{v}_{\mathrm{P}}=+30 \mathrm{~cm} $$
The image formed by lens P is 30 cm to the right of lens P .
So, the magnification of image formed by lens P is,
$$ \mathrm{m}_{\mathrm{P}}=\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{u}_{\mathrm{P}}}=\frac{30}{-15}=-2 $$
Lens Q is placed 15 cm to the right of lens P and the image from lens P acts as the object for lens Q .
Since the image from P is at 30 cm and lens Q is at 15 cm , the object for lens Q is $30-15=15 \mathrm{~cm}$ to the right of lens Q.
The image formed by P acts as virtual object for the lens Q .
For lens Q :
Object distance is $\mathrm{u}_{\mathrm{Q}}=+15 \mathrm{~cm}$
Focal length is $\mathrm{f}_{\mathrm{Q}}=+15 \mathrm{~cm}$
Using the lens formula,
$$ \frac{1}{v_Q}-\frac{1}{u_Q}=\frac{1}{f_Q} $$
$$\Rightarrow $$ $\frac{1}{v_Q}-\frac{1}{15}=\frac{1}{15}$
$$\Rightarrow $$ $\frac{1}{v_Q}=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}$
$$\Rightarrow $$ $$ \mathrm{v}_{\mathrm{Q}}=+7.5 \mathrm{~cm} $$
If the object is real and the image is formed on opposite side of the convex lens, then image so formed is real but if the object is virtual and image is formed on the same side of the convex lens, then the image so formed is real.
The final image formed by lens Q is real and formed 7.5 cm to the right of lens Q .
So, the magnification of image formed by lens P is,
$$ \mathrm{m}_{\mathrm{Q}}=\frac{\mathrm{v}_{\mathrm{Q}}}{\mathrm{u}_{\mathrm{Q}}}=\frac{7.5}{15}=0.5 $$
Total magnification $\mathrm{M}=\mathrm{m}_{\mathrm{P}} 	imes \mathrm{m}_{\mathrm{Q}}$
$$ M=(-2) 	imes(0.5)=-1 $$
Since $|\mathrm{M}|=1$, the size of the final image is the same as that of object AB .
Therefore, the final image is real, formed at 7.5 cm right of lens Q , with a size same as that of AB .
Thus, the correct option is (B).

Answer

virtual, formed at 7.5 cm right of lens $Q$, with a size bigger than that of $A B$ real, formed at 7.5 cm right of lens

Answer

$Q$, with a size same as that of $A B$

Answer

formed at infinity.

Answer

real, formed at 7 cm right of lens $Q$, with a size smaller than that of $A B$

Ray Optics

An object A B is placed 15 cm on the left of a convex lens P of focal length 10 cm . Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm , the final image is \_\_\_\_ .

Choose the correct answer: