Question

1.
An object is placed at a distance 20 cm from
a convex lens of focal length 10 cm. Find the
image distance and nature of the image.​

Answer 1

Answer:By using lens formula - $\\ \\ \sf \: \: \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} \\ \\ \\ \implies \sf \: \frac{1}{v} = \frac{1}{20} - \frac{1}{ - 10} \\ \\ \\ \implies \sf \: \frac{1}{v} = \frac{3}{20} \\ \\ \\ \implies \sf \: v = 6.67 \: cm \\ \\ \\ \large \underline{ \boxed{ \sf{ \green{v = 6.67}}}}$Therefore, the image is formed at a distance of 6.67 cm .Hence, the sign of the image is positive it means it indicates the nature of the image is virtual and errect.