Question

A 2.7-cm-tall object is 20 cm to the left of a lens with a focal length of 10 cm . A second lens with a focal length of 48 cm is 52 cm to the right of the first lens.

A. Calculate the distance between the image and the second lens.B.Calculate the image height.

Answer 1

Answer

given,

height of object = 2.7 cm

distance left of lens (u₁)= 20 cm

focal length of lens(f₁)= 10 cm  

the distance of image

$\dfrac{1}{f_1}=\dfrac{1}{u_1}+\dfrac{1}{v_1}$

$\dfrac{1}{v_1}=\dfrac{1}{f_1}-\dfrac{1}{u_1}$

$\dfrac{1}{v_1}=\dfrac{1}{10}-\dfrac{1}{20}$

   v₁ = 20 cm

magnification of first lens

$m_1= -\dfrac{v}{u}$

$m_1=-\dfrac{20}{20}$

    m₁ = -1

distance of object from the second lens

   u₂ = 52-20 = 32 cm

   f₂ = 48 cm

now,

$\dfrac{1}{f_2}=\dfrac{1}{u_2}+\dfrac{1}{v_2}$

$\dfrac{1}{v_2}=\dfrac{1}{f_2}-\dfrac{1}{u_2}$

$\dfrac{1}{v_2}=\dfrac{1}{48}-\dfrac{1}{52}$

   v₁ = 624 cm

magnification of first lens

$m_1= -\dfrac{v}{u}$

$m_1=-\dfrac{624}{52}$

    m₁ = -12

total magnification

m = m₁ m₂

m = (-1)(-12)

m = 12

height of image

$m =-\dfrac{h'}{h}$

$12=-\dfrac{h'}{2.7}$

h' = -32.4 cm

a) distance between image and second lens is equal to 624 cm

b) height of image is equal to 32.4 cm

       

Answer 2

(a) The distance between the image and the second lens is 34 cm.

(b) The height of the image formed is 2.7 cm.

Image distance from the first lens

The image distance from the first lens is calculated as follows;

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\\\\frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\\\\frac{1}{v} =\frac{1}{10} - \frac{1}{20} \\\\\frac{1}{v} = \frac{1}{20} \\\\v = 20 \ cm$

Magnification of the image is 1, same size as the object.

Distance between the image and the second lens

Distance = 54 cm - 20 cm = 34 cm

Since the magnification is 1, the image height is equal to object height = 2.7 cm.

Learn more about image formed by lens here: https://brainly.com/question/25736513