To solve this problem, we need to analyze the optical system consisting of an equiconvex lens and a concave mirror. We will use the lens formula and the mirror formula to find the distance \( d \) between the lens and the mirror. Let's break down the steps clearly.
Understanding the System
We have an equiconvex lens with a focal length \( f = 20 \, \text{cm} \) and a concave mirror with a radius of curvature \( R = 80 \, \text{cm} \). The focal length \( f_m \) of the concave mirror can be calculated using the formula:
- Focal length of the mirror: \( f_m = -\frac{R}{2} = -\frac{80}{2} = -40 \, \text{cm} \)
The negative sign indicates that the focal point of the concave mirror is on the same side as the object.
Position of the Object
The point object \( O \) is placed at a distance of \( 40 \, \text{cm} \) from the lens. We need to find the image formed by the lens first, as this image will serve as the object for the concave mirror.
Using the Lens Formula
The lens formula is given by:
- Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Where:
- \( f \) is the focal length of the lens (20 cm)
- \( v \) is the image distance from the lens
- \( u \) is the object distance from the lens (which is -40 cm, since we take the object distance as negative in lens convention)
Substituting the values into the lens formula:
\( \frac{1}{20} = \frac{1}{v} - \frac{1}{-40} \)
Rearranging gives:
\( \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \)
Finding a common denominator (40):
\( \frac{1}{v} = \frac{2}{40} - \frac{1}{40} = \frac{1}{40} \)
Thus, \( v = 40 \, \text{cm} \). This means the image formed by the lens is located 40 cm on the opposite side of the lens.
Image as Object for the Mirror
Now, this image acts as the object for the concave mirror. Since the image is formed 40 cm from the lens, we need to find the distance \( d \) from the lens to the mirror. The total distance from the object \( O \) to the mirror is \( d + 40 \, \text{cm} \).
Applying the Mirror Formula
The mirror formula is:
- Mirror formula: \( \frac{1}{f_m} = \frac{1}{v_m} + \frac{1}{u_m} \)
Here, \( u_m \) is the object distance for the mirror, which is negative because the object (the image from the lens) is on the same side as the incoming light. Thus, we have:
\( u_m = -(d + 40) \)
And the image distance \( v_m \) for the mirror is positive since the final image is formed at the position of the object \( O \), which is 40 cm from the mirror:
\( v_m = 40 \, \text{cm} \)
Substituting these values into the mirror formula:
\( \frac{1}{-40} = \frac{1}{40} + \frac{1}{-(d + 40)} \)
Rearranging gives:
\( \frac{1}{-(d + 40)} = \frac{1}{-40} - \frac{1}{40} \)
Which simplifies to:
\( \frac{1}{-(d + 40)} = 0 \)
This indicates that the object distance \( d + 40 \) must equal 0, leading us to find:
\( d + 40 = -40 \)
Thus, solving for \( d \):
\( d = -80 \, \text{cm} \)
Final Result
The distance \( d \) between the lens and the concave mirror is \( 80 \, \text{cm} \) on the side of the lens where the object is placed. This means the mirror is positioned 80 cm away from the lens, allowing the system to form the final image at the position of the original object \( O \).
