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