Question

an object with the height of 6 cm is placed in the front of concave lens of focal length 20 cm is placed at a distance of 30cm find nature size and position the image and also draw the diagram​

Answer 1

Answer:

Given : u=-30 cm f= -20 cm h =6cm

Using lens formula :

$\frac{1}{f } = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{ - 20 } = \frac{1}{v} - \frac{1}{ - 30}$

$\frac{1}{ - 20 } = \frac{1}{v} + \frac{1}{ 30}$

$\frac{1}{ v } = - \frac{1}{ 20} - \frac{1}{ 30}$

$\frac{1}{ v } = \frac{ - 3 - 2}{60}$

$v = \frac{60}{ - 5}$

$v = - 12cm$

(- sign shows that image is virtual)

$m = \frac{h {i} }{ho} = \frac{v}{u}$

$\frac{h {i} }{6} = \frac{ - 12 }{ - 30}$

hi = +12/5 (+ sign shows that image is erect)

Answer 2

Answer:

Hope the above image helps you