Question

5cm high object is placed at a distance of 25cm from converging lens of focal length of 10 cm. determine the position size and type of the image.​

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,

$\bold{= \frac{1}{f}=\frac{1}{v}-\frac{1}{u}}\\ \\ \bold{= \frac{1}{10}=\frac{1}{v}-\frac{1}{-25}}$

⇒   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.