To solve this problem, we need to analyze the optical system consisting of two lenses, one of which is silvered. The silvered lens acts as a concave mirror due to the reflective surface. Let’s break down the steps to find the height of the object above the upper lens.
Understanding the System
We have two lenses in this setup:
- The first lens (convexo-convex) has a focal length \( f_1 = 40 \, \text{cm} \) and is silvered on one side, effectively making it a mirror.
- The second lens (also convex) has a focal length \( f_2 = 20 \, \text{cm} \) and is positioned 10 cm above the first lens.
Calculating the Effective Focal Length of the Silvered Lens
For the first lens, we can find the radius of curvature \( R_1 \) using the lens maker's formula:
Lens Maker's Formula: \(\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\)
Since one side is silvered, we can consider it as a mirror with \( R_2 = -R_1 \). Given \( R_2 = 120 \, \text{cm} \) (positive for the convex surface), we have:
Let \( R_1 = 120 \, \text{cm} \) and \( R_2 = -120 \, \text{cm} \). Plugging these values into the formula:
\(\frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{120} - \frac{1}{-120} \right)\)
\(\frac{1}{f_1} = 0.5 \left( \frac{2}{120} \right) = \frac{1}{120} \Rightarrow f_1 = 40 \, \text{cm}\)
Finding the Image from the First Lens
Now, we need to find the image formed by the first lens (acting as a mirror). The object distance \( u_1 \) is negative (since it is on the same side as the incoming light). Let’s denote the object distance as \( u_1 = -h \) (where \( h \) is the height of the object above the upper lens).
Using the mirror formula:
Mirror Formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
Rearranging gives us:
\(\frac{1}{v_1} = \frac{1}{f_1} - \frac{1}{u_1}\)
Substituting \( f_1 = 40 \, \text{cm} \) and \( u_1 = -h \):
\(\frac{1}{v_1} = \frac{1}{40} + \frac{1}{h}\)
Image Distance from the First Lens
Now, we need to find the distance of the image formed by the first lens, which will act as the object for the second lens. The distance from the first lens to the second lens is 10 cm, so:
\( v_1 = h + 10 \)
Substituting this into the equation gives:
\(\frac{1}{h + 10} = \frac{1}{40} + \frac{1}{h}\)
Solving for the Object Height
Cross-multiplying to eliminate the fractions leads to:
\( h + 10 = \frac{h \cdot 40}{h + 40} \)
Multiplying both sides by \( (h + 40) \) results in:
\( (h + 10)(h + 40) = 40h \)
Expanding gives:
\( h^2 + 50h + 400 = 40h \)
Rearranging leads to:
\( h^2 + 10h + 400 = 0 \)
Using the Quadratic Formula
Now we can apply the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 10, c = 400 \):
Calculating the discriminant:
\( b^2 - 4ac = 10^2 - 4 \cdot 1 \cdot 400 = 100 - 1600 = -1500 \)
Since the discriminant is negative, this indicates that there are no real solutions for \( h \). This suggests that the configuration of the lenses and the object position may not allow for a real image coincident with the object.
Final Thoughts
In this scenario, it appears that the object cannot be positioned such that the image coincides with it given the parameters of the lenses. This could be due to the specific focal lengths and distances involved. If you have any further questions or need clarification on any part of the process, feel free to ask!
