Respuesta :
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
[tex]\dfrac{1}{f_1}=\dfrac{1}{u_1}+\dfrac{1}{v_1}[/tex]
[tex]\dfrac{1}{v_1}=\dfrac{1}{f_1}-\dfrac{1}{u_1}[/tex]
[tex]\dfrac{1}{v_1}=\dfrac{1}{10}-\dfrac{1}{20}[/tex]
v₁ = 20 cm
magnification of first lens
[tex]m_1= -\dfrac{v}{u}[/tex]
[tex]m_1=-\dfrac{20}{20}[/tex]
m₁ = -1
distance of object from the second lens
u₂ = 52-20 = 32 cm
f₂ = 48 cm
now,
[tex]\dfrac{1}{f_2}=\dfrac{1}{u_2}+\dfrac{1}{v_2}[/tex]
[tex]\dfrac{1}{v_2}=\dfrac{1}{f_2}-\dfrac{1}{u_2}[/tex]
[tex]\dfrac{1}{v_2}=\dfrac{1}{48}-\dfrac{1}{52}[/tex]
v₁ = 624 cm
magnification of first lens
[tex]m_1= -\dfrac{v}{u}[/tex]
[tex]m_1=-\dfrac{624}{52}[/tex]
m₁ = -12
total magnification
m = m₁ m₂
m = (-1)(-12)
m = 12
height of image
[tex]m =-\dfrac{h'}{h}[/tex]
[tex]12=-\dfrac{h'}{2.7}[/tex]
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
(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;
[tex]\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[/tex]
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