Answer to Question #113469 in Optics for Hayley

Question

An object is placed 20.0 cm from a converging lens with focal length 15.0 cm (see the figure, not drawn to scale). A concave mirror with focal length 10.0 cm is located 78.5 cm to the right of the lens. Light goes through the lens, reflects from the mirror, and passes through the lens again, forming a final image.
What is the overall transverse magnification?

Answer 1

As per the given question,

Distance of object $(s_1)=20.0cm$

focal length of the conversing lens $(f_1)=15.0cm$

focal length of the concave mirror $(f_2)=10cm$

distance between the lens and mirror $=78.5cm$

Let the position of image for the lens is $s_1'$

So,

$s_1'=\dfrac{s_1f_1}{s_1-f_1}$

$\Rightarrow s_1'=\dfrac{15\times 20}{20-15}=60cm$

Now, this image work as object for the concave mirror,

the distance between the first image by convex lens and mirror$(s_2)=78.5-60=18.5cm$

Now, let the image position is $s_2'$

$s_2'=\dfrac{s_2f_2}{f_2-s_2}=\dfrac{18.5\times 10}{-8.5}=-21.76cm$

overall transverse magnification $(m)=\dfrac{21.76}{20}=1.088$