Question

The light rays are incident on a convex lens as in figure. If focal length of lens is 20 cm then find the position of image. Show ray diagram as well. 

 

Hint:

Remember that a converging beam incident on a lens or mirror corresponds to a virtual object, and its distance (u) is taken as positive according to the Cartesian sign convention.

Answer 1

Step 1: Understanding the Concept: 
The given figure shows a beam of light that is already converging towards a point O. When this beam passes through the convex lens, the point O acts as a virtual object. We need to use the lens formula to find the position of the final image. 
Step 2: Key Formula or Approach: 
We will use the lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] And the standard sign convention:
Light travels from left to right.
Distances measured in the direction of light are positive.
Distances measured opposite to the direction of light are negative.
The focal length (f) of a convex lens is positive.
Step 3: Detailed Explanation: 
From the problem and the diagram:
Focal length (f): The lens is convex, so \( f = +20 \) cm.
Object distance (u): The incident rays are converging to a point O which is 10 cm to the right of the lens. Since this point is where the rays would meet if the lens were not present, it acts as a virtual object. Its distance is measured from the optical center in the direction of light, so it is positive. \( u = +10 \) cm.
Now, we substitute these values into the lens formula to find the image distance (v): \[ \frac{1}{v} - \frac{1}{+10} = \frac{1}{+20} \] \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{10} \] \[ \frac{1}{v} = \frac{1 + 2}{20} = \frac{3}{20} \] \[ v = \frac{20}{3} \approx +6.67 \, \text{cm} \] The positive sign for v indicates that the image is formed on the right side of the lens. Since the light rays actually converge at this point, the image is real. 
Step 4: Ray Diagram: 

The diagram shows the converging incident rays (dashed) heading towards the virtual object O. The convex lens converges them further, causing them to meet at the real image I, which is formed closer to the lens.