Solveeit Logo
Login

Question

Question: An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. On the other si...

An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed such that its distance from the lens equals focal length of the lens. The focal length of the convex mirror is?
A. 20 cm
B. 10 cm
C. 15 cm
D. 30 cm

Explanation

Solution

Convex lens is a converging lens and its focal length is positive. The object is placed at the left and taking sign convention it will be negative. We need to find the focal length of the convex mirror. Convex mirror, we know it always forms a virtual image. Virtual image is the image in which rays appear to meet.

Complete step by step solution:
FOR CONVEX LENS;
u= object distance= -15 cm
focal length of lens= f= 10 cm
image distance, v=?
using lens formula,

1f=1v−1u ⇒1f+1u=1v ⇒1v=110−115 ⇒1v=3−230 ⇒1v=130 ∴v=30cm \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{v} \\\ \Rightarrow \dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{15} \\\ \Rightarrow \dfrac{1}{v}=\dfrac{3-2}{30} \\\ \Rightarrow\dfrac{1}{v}=\dfrac{1}{30} \\\ \therefore v=30cm \\\

So, the image is formed at a distance of 30 cm to the right side of the lens. It will act as an object for the mirror. Since, the image distance comes out to be positive, so as per sign convention, it is on the right side of the mirror.

FOR CONVEX MIRROR;
Distance between the mirror and the lens is 10 cm, so the object for convex mirror is at a distance of (30-10) = 20 cm
u= 20 cm
so focal length= 202=10cm\dfrac{20}{2}=10cm

Therefore, the correct option is (B)

Additional Information: Magnification if greater than one then the image is enlarged and if it is less than one then image is diminished. If the magnification is negative then image is inverted and if magnification is positive then image is erect. If magnification is unity then size of the image is equal to the size of the object.

Note: While doing such problems we should keep in our mind that if the object is placed at the left side of the mirror or the lens, the distance will always be taken in negative. We follow cartesian sign conventions. All distances above the principal axis are taken as positive and all below the principal axis are taken as negative.