Question

The distance of object needle is 45 cm from a lens which forms an image on the screen placed 90 cm on the other side. What is the type of lens? What is the focal length and the size of the image if the size of object needle is 5 cm?

Hint:

The magnification \( m \) indicates the relative size and orientation of the image. A negative value indicates an inverted image.

Answer 1

Step 1: Identify the type of lens. 
The image formed is on the opposite side of the object, indicating that the lens is a **converging lens** (a convex lens). 
Step 2: Use the lens formula. 
The lens formula relates the object distance \( u \), image distance \( v \), and focal length \( f \) of a lens: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where: - \( u = -45 \, \text{cm} \) (object distance, negative since it is on the left side of the lens), - \( v = 90 \, \text{cm} \) (image distance, positive since it is on the right side of the lens). 
Substitute the values into the lens formula: \[ \frac{1}{f} = \frac{1}{90} - \frac{1}{-45} \] \[ \frac{1}{f} = \frac{1}{90} + \frac{1}{45} \] \[ \frac{1}{f} = \frac{1 + 2}{90} = \frac{3}{90} = \frac{1}{30} \] So, the focal length \( f = 30 \, \text{cm} \). 
Step 3: Calculate the size of the image. 
The magnification \( m \) of a lens is given by the formula: \[ m = \frac{\text{Image height}}{\text{Object height}} = \frac{v}{u} \] The object size is given as 5 cm, so: \[ m = \frac{v}{u} = \frac{90}{-45} = -2 \] This means the image is inverted (negative magnification) and is twice the size of the object. 
Thus, the size of the image is: \[ \text{Image size} = |m| \times \text{Object size} = 2 \times 5 = 10 \, \text{cm} \] Step 4: Conclusion. 
- The lens is a **convex lens** (converging lens). - The **focal length** of the lens is \( 30 \, \text{cm} \). - The **size of the image** is \( 10 \, \text{cm} \).