Refraction At Spherical Surfaces And Lenses

The thin-lens equation from NCERT connects the focal length to the object and image distances through $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, with the Cartesian convention giving the object distance a negative value. Here the object is 40 cm to the left, so $u=-40$ cm, and the image forms on a screen 40 cm to the right, so $v=+40$ cm. Substituting, $\frac{1}{f}=\frac{1}{40}-\frac{1}{-40}=\frac{1}{40}+\frac{1}{40}=\frac{2}{40}=\frac{1}{20}$, hence $f=20$ cm. This is the special symmetric case where object and image sit at twice the focal length, producing a real, inverted, same-size image. The value 40 cm is wrong because it mistakes the equal object and image distances for the focal length itself. The value 80 cm is wrong because it adds instead of taking the reciprocal sum. The value 10 cm is wrong because it doubles the reciprocal incorrectly. As stated in NCERT Class 12, Chapter 9, when the object is at $2f$ the image also forms at $2f$ with unit magnification. A sanity check confirms that placing the object at $2f=40$ cm indeed yields a screen image at 40 cm, exactly the described situation.