Question

 A 4cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20cm. If the distance of the object is 30cm from the lens, find the position, size of the image. Also find its magnification.​

Answer 1

Answer:

your answer dear........see attachment

Answer 2

Given,

• Object Distance (u) = -30 cm

• Focal Length (f) = 20 cm

• Object Height ($\sf h_o$) = 4 cm

We know the Lens' Formula,

$\boxed{\bf \dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u}}$    

$\boxed{\bf \dfrac{1}{v} =\dfrac{1}{f} +\dfrac{1}{u}}$

Applying the values to the equation,

$\implies\sf \dfrac{1}{v}=\dfrac{1}{20} + \dfrac{-1}{30}$

$\implies\sf \dfrac{1}{v}=\dfrac{1}{20} - \dfrac{1}{30}$

$\implies\sf \dfrac{1}{v}=\dfrac{30}{600} - \dfrac{20}{600}$

$\implies\sf\dfrac{1}{v}= \dfrac{30-20}{600}$

$\implies\sf\dfrac{1}{v}= \dfrac{10}{600}$

$\implies\sf\dfrac{1}{v}= \dfrac{1}{60}$

$\implies\sf v=\dfrac{60}{1}$

$\implies\sf v=60\:cm$

So,

$\boxed{\bf Position\:of\:image=60\:cm}$

We know the Magnification equation,

$\boxed{\bf Magnification\:(M)= \dfrac{Image \:Distance\:(u)}{Object \:Distance\:(v)}}$

Applying the values to the equation,

$\implies\sf M=\dfrac{60}{-30}$

$\implies\sf M=-2$

So,

$\boxed{\bf Magnification=-2\:M}$

Now, we have to find the image height from the magnification equation,

$\boxed{\bf Magnification\:(M)= \dfrac{Image \:Height\:(h_i)}{Object \:Height\:(h_o)}}$

Applying the values to the equation,

$\implies\sf -2=\dfrac{h_i}{4}$

$\implies\sf h_i=-2\times4$

$\implies\sf h_i=-8\:cm$

Height cannot be negative.

$\implies\sf h_i=8\:cm$

$\boxed{\bf Height\:of\:image=8\:cm}$

Therefore,

• Position of Image = 60 cm

• Height of Image = 8 cm

• Magnification = -2 M

• Nature = Real + Inverted