Question:
A diverging lens, with a focal length of 15 cm, is placed 30 cm in front of a concave mirror with a focal length of 20 cm. A parallel beam of light, consisting of rays, is incident on the lens and hits the mirror. The light rays emerge from the mirror and converge to a point 50 cm from the lens. Determine the final position and nature (real/virtual) of the image formed by this optical system. Show all calculations and steps.
Answer:
Given: Focal length of diverging lens, f1 = -15 cm (negative for divergence) Focal length of concave mirror, f2 = -20 cm (negative for concavity) Distance of the lens from the object, do = 30 cm Distance of the final image from the lens, di = 50 cm
To solve this problem, we need to use the lens and mirror formula, which are:
Lens formula: 1/f = 1/do + 1/di Mirror formula: 1/f = 1/do + 1/di
Substituting the given values into the lens formula, we have:
1/(-15) = 1/30 + 1/di
Simplifying the equation, we get:
-1/15 = 1/30 + 1/di
Now, let's solve for di in terms of a common denominator:
-1/15 = (2 + di)/60
To eliminate the fraction, we cross-multiply:
-60 = 2 + di
Subtracting 2 from both sides of the equation:
di = -62 cm
So, the final image is formed 62 cm in front of the lens. Since the object is placed in front of the lens, and the image is formed on the same side as the object, the image is virtual.
Now, let's analyze the mirror equation. We will use the same values as before:
1/(-20) = 1/30 + 1/di
Simplifying the equation, we get:
-1/20 = 1/30 + 1/di
Next, let's solve for di in terms of a common denominator:
-1/20 = (2 + di)/60
Cross-multiplying to eliminate the fraction:
-60 = 2 + di
Subtracting 2 from both sides of the equation:
di = -62 cm
The final image formed by the mirror is also 62 cm in front of it. Since the image is formed in front of the mirror, it is considered real.
To summarize: