To determine the focal length (f) and position (s) of the lens needed to create a double-sized image of a spider on a wall, we can use the lens formula and magnification concepts.
Given data:
- Diameter of the spider = 20 cm (which translates to the height of the spider, h = 20 cm)
- Distance from the spider to the wall = 2.6 m = 260 cm
- Desired magnification (M) = 2
1. **Calculating the Position (s) of the Lens**:
The magnification (M) of a lens is given by:
M = – (Image distance (v) / Object distance (u))
We set the object distance (u) as the distance from the lens to the spider. The total distance from the lens to the wall is 260 cm. So, if we denote the distance from the lens to the spider as u, the image distance will be:
v = 260 – u
Substituting into the magnification formula:
2 = – ((260 – u) / u)
Cross multiplying gives:
2u = – (260 – u)
So:
2u = -260 + u
Rearranging terms:
u + 260 = 0
This simplifies to:
u = 86.67 cm (approx)
2. **Calculating the Image Distance (v)**:
Using the found value of u:
v = 260 – 86.67 = 173.33 cm (approx)
3. **Finding the Focal Length (f)**:
The lens formula states:
rac{1}{f} = rac{1}{u} + rac{1}{v}
Substituting the values of u and v:
rac{1}{f} = rac{1}{86.67} + rac{1}{173.33}
Calculating these values gives:
rac{1}{f} = 0.0115 + 0.0058
ightarrow rac{1}{f} = 0.0173
Finally, we find f:
f ext{ (focal length)} = rac{1}{0.0173}
ightarrow f ≈ 57.8 cm (approx)
4. **Conclusion**:
To achieve a double-sized image of the 20 cm diameter spider on the wall that is 2.6 m away, the focal length of the lens should be approximately 57.8 cm, and it should be placed around 86.67 cm from the spider.