Given: An equiconvex lens,
f1=10cm, is placed
40cm in front of a concave mirror,
f2=7.50cm. An object
2cm high is placed
20cm to the left of the lens. If overall magnification is
X1000.
To find the value of X
Solution:
From lens formula
1v−1−20=110⟹v=+20cm
Therefore magnification of first image, m1=vu=−1
The image formed by the equiconvex lens right will be real, inverted and of same size as the object.
The first image will act as the object for concave mirror. Object distance for mirror is (40−20)cm.
From mirror equation,
1v′+1−20=1−7.5⟹v′=−12cm
Therefore the second magnification will be, m2=v′u′=−12−20=0.6
The second image acts as object for the lens. The object distance for the second refraction at the lens, u′′=+28cm
From lens equation,
1v′′−1+28=1−10⟹v′′=−15.6cm
So final magnification m3=v′′u′′=−15.6+28=−0.556
Therefore, overall magnification will be
m=m1×m2×m3⟹m=(−1)(0.6)(−0.556)=−0.333⟹m=3331000
Hence X=333