Question

5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image. (Ans : 16.7 cm, 3.3 cm, Real) Solve the example

Answer 1

Given conditions ⇒
Height of the Object(H₀) = 5 cm. 
Object Distance(u) = 25 cm. 
Focal length(f) = 10 cm.

Since, the lens is converging, 
∴ u = -25 cm.
f = 10 cm. 

Using the lens formula, 
    $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
 ⇒ $\frac{1}{10} = \frac{1}{v} - \frac{1}{-25}$
⇒   v = 50/3 
∴ v = 16.67 cm. 

Since, the image distance is positive. This means that the real image is formed by the Lens. Also, It is formed on the Opposite size of the lens. I means on the side where the x-axis is present. 

Now, Magnification = v/u
   = -16.67/25
   = - 0.6668 

Also, Magnification = H₁/H₀
∴ H₁ = 0.6668 × 5
        = 3.334 cm.
        ≈ 3.3 cm. 


Since, the height of the image is less than than that of the object, therefore the image is diminished with respect to the object. 


Hope it helps.

Answer 2

This answer will help you