Question

Q6. An object is held 30 cm away from a convex lens of focal length 10
cm. What is the position and nature of the image formed.​

Answer 1

Answer:

the image is formed between center of curvature and focal point and image is small and invert

Answer 2

By sign convention ,

• Object distance (u) = -30 cm
• Focal length (f) = 10 cm

We know that , the lens formula is given by

$\huge \mathtt{ \fbox{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }}$

Thus ,

$\sf \mapsto \frac{1}{10} = \frac{1}{v} - ( - \frac{1}{30} ) \\ \\ \sf \mapsto \frac{1}{v} = \frac{1}{10} - \frac{1}{30} \\ \\ \sf \mapsto \frac{1}{v} = \frac{ 30 - 10}{300} \\ \\ \sf \mapsto v = \frac{300}{20} \\ \\ \sf \mapsto v = 15 \: \: cm$

Positive sign of v indicates the image is virtual

Hence , the virtual image is formed 7.5 cm from the left side of the lens