Question

A biconvex lens has a focal length of 12.0 cm. If an object is placed 6.00 cm from the lens, compute the image distance and characteristics including magnification. Indicate if the image is real or virtual. Be careful with the signs (+ or -) for this problem.​

Answer 1

Given:

A biconvex lens has a focal length of 12.0 cm.

To find:

• Image distance
• Image characteristics
• Magnification

Calculation:

Applying Len's Formula:

$\rm \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\rm \implies \dfrac{1}{12} = \dfrac{1}{v} - \dfrac{1}{( - 6)}$

$\rm \implies \dfrac{1}{12} = \dfrac{1}{v} + \dfrac{1}{6}$

$\rm \implies \dfrac{1}{v} = \dfrac{1}{12} - \dfrac{1}{6}$

$\rm \implies \dfrac{1}{v} = \dfrac{1 - 2}{12}$

$\rm \implies \dfrac{1}{v} = \dfrac{ - 1}{12}$

$\rm \implies v = - 12 \: cm$

• Image distance is -12 cm, i.e. before the lens.

• So, image is virtual, erect and magnified.

$\rm mag. = \dfrac{v}{u}$

$\rm \implies mag. = \dfrac{ - 12}{ - 6}$

$\rm \implies mag. = 2$

So, magnification is 2.