Answer:
the required distance is 6 cm
Explanation:
Given the data in the question;
f₁ = 15 cm
f₂ = 5.0 cm
d = 45 cm
Now, for first lens object distance s = ∝
1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'
Now, image distance of first lens s' = 15cm
object distance of second lens s₂ will be;
s₂ = 45 - 15 = 30 cm
so
1/f₂ = 1/s₂ + 1/s'₂
1/5 = 1/30 + 1/s'₂
1/s'₂ = 1/5 - 1/30
1/s'₂ = 1 / 6
s'₂ = 6 cm
Hence, the required distance is 6 cm
The distance of the final image from the first lens will be is 6 cm.
The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).
The given data in the problem is;
f₁ is the focal length of lens 1= 15 cm
f₂ s the focal length of lens 2= 5.0 cm
d is the distance between the lenses = 45 cm
From the mirror equation;
[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]
If f₁ is the focal length of lens 1 is 15 cm then;
[tex]s'=15 cm[/tex]
f₂ s the focal length of lens 2= 5.0 cm
s₂ = 45 - 15 = 30 cm
From the mirror equation;
[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]
Hence the distance of the final image from the first lens will be is 6 cm.
To learn more about the mirror equation refer to the link;
https://brainly.com/question/3229491
