Solved Problems on Lens Sign Convention
Example 1: An object is placed 20 cm from a converging lens with a focal length of 10 cm. Calculate the image distance and determine the nature (real or virtual), orientation (upright or inverted), and size of the image formed.
Solution:
Given:
- Object distance, u = 20 cm (positive, as it’s in front of the lens)
- Focal length, f = 10 cm (positive, as it’s for a convex lens)
Using the lens equation:
1/v + 1/u = 1/f
Substituting the given values:
1/v + 1/20 = 1/10
Now, let’s solve for v :
1/v = 1/10 – 1/20 = 1/20
So, the image distance is v = 20 cm. Now, let’s analyze the characteristics of the image:
Since the image distance (20 cm) is positive, the image is formed on the opposite side of the lens from the object, indicating it is a real image.
To determine if the image is upright or inverted, we need to find the magnification.
The magnification m is given by the formula: m=-v/u
Substituting the given values: m=-1
Since the magnification is negative, the image is inverted relative to the object and the size of the image is the same as the size of the object.
Therefore, the image formed by the lens is a real image, inverted, and the same size as the object.
Example 2: An object is placed 15 cm from a converging lens. The image formed is real, inverted, and four times the size of the object. Calculate the focal length of the lens.
Solution:
Given:
- Object distance, u = 15 cm (positive, as it’s in front of the lens)
- Magnification, m = -4 (negative because inverted image)
The magnification m is given by the formula: m=-v/u
Now, let’s solve for v :
-4 = -v/15
So, the image distance is v = 60 cm.
Now that we have both v and u, using the lens equation:
1/v + 1/u = 1/f
Substituting the given values:
1/f = 1/60 + 1/15 = 3/60 = 1/20
So, the focal length is f = 20 cm.
Example 3: A 4.00-cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of -12.2 cm. Determine the image distance and the image size.
Solution:
Given:
- Object distance, u = 35.5 cm
- Focal length, f = -12.2 cm (negative, as it’s for a concave lens)
Using the lens equation:
1/v + 1/u = 1/f
Substituting the given values:
1/v + 1/35.5 = 1/-12.2
Now, let’s solve for v:
1/v = 1/-12.2 – 1/35.5
⇒ 1/v = – [(35.5+12.2)/433.1]
⇒ 1/v = -(47.7/433.1)
⇒ 1/v =- 0.110
⇒ v = – (1/0.110)
⇒ v = -9.09 cm
So, the image distance is v=-9.09 cm.
The magnification m is given by the formula: m=-v/u
Substituting the given values: m=-(-9.09)/35.5=0.256
Since the magnification is positive, the image is upright relative to the object.
The height of image , hi=m x ho=0.256 x 4=1.02 cm
Therefore, the image formed by the lens is a virtual image, erect, and the size is 1.02 cm.
Lens Sign Convention
A lens in optics is a transparent device with curved surfaces that refract light. It can be converging or diverging based on its shape. The lens sign convention, or Cartesian sign convention, helps determine the nature, size, and position of images formed by lenses accurately. It clarifies the positive and negative signs for object and image distances, focal lengths, and magnifications, crucial for optical calculations and predictions.
Table of Content
- What is the Lens Sign Convention?
- Basics of Lens Sign Convention
- Cartesian Sign Convention
- Sign Conventions in Different Lenses
- Lens Maker’s formula
- Conclusion: Lens Sign Convention