An optical projector is used to project an image of a slide 50mm square on to a screen places 1.2 m from the lens. The image formed on the screen is 600mm square. Find the focal length of the projecting lens of the projector.

Note: convert all unit measurements to cm(centimeter)

1 answer

To find the focal length of the projecting lens, we can use the lens formula and the magnification formula.

  1. Convert all measurements to centimeters:

    • Size of the slide: \( 50 , \text{mm} = 5 , \text{cm} \)
    • Size of the image on the screen: \( 600 , \text{mm} = 60 , \text{cm} \)
    • Distance from the lens to the screen: \( 1.2 , \text{m} = 120 , \text{cm} \)
  2. Calculate the magnification: \[ \text{Magnification} (M) = \frac{\text{Image size}}{\text{Object size}} = \frac{60 , \text{cm}}{5 , \text{cm}} = 12 \]

  3. Using the magnification formula: Magnification is also given by: \[ M = -\frac{v}{u} \] where:

    • \( v \) is the image distance (120 cm),
    • \( u \) is the object distance.

    We can rearrange this to find \( u \): \[ 12 = -\frac{120}{u} \implies u = -\frac{120}{12} = -10 , \text{cm} \]

    The negative sign indicates that the object (the slide) is on the same side of the lens as the incoming light (which is conventional).

  4. Using the lens formula: The lens formula relates object distance, image distance, and the focal length: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Plugging in the values: \[ \frac{1}{f} = \frac{1}{120} + \frac{1}{-10} \] Find a common denominator (120): \[ \frac{1}{f} = \frac{1}{120} - \frac{12}{120} = \frac{1 - 12}{120} = \frac{-11}{120} \] Therefore: \[ f = -\frac{120}{11} \approx -10.91 , \text{cm} \]

However, in lens optics, we typically consider the focal length to be positive for a converging lens (which is usually the case for projectors). As it’s a projector lens, we take the absolute value:

Thus, the focal length of the projecting lens is approximately: \[ \boxed{10.91 , \text{cm}} \]