Question

01. A concave mirror has a radius of curvature of 0.3m. An object 4m in size, is placed at 25m infront of the mirror. Find size of the image​

Answer 1

Explanation:

Hello Dear.

Here is the answer---

Height of the Object(H₀) = 4 cm.

Object Distance(u) = 25 cm.(negative)

Focal Length = 15 cm.(negative)

Using the Mirror's Formula,

    \frac{1}{f} = \frac{1}{v} + \frac{1}{u}f1=v1+u1

⇒  \frac{1}{-15} = \frac{1}{-25} + \frac{1}{v}−151=−251+v1

⇒  v = - 37.5 cm.

Thus, the image is formed behind the mirror. Since the Distance is Negative, thus the Image is Real.

Thus, Screen is to be placed at a Distance of 37.5 cm behind the Mirror.

Now, Magnification = H₁/H₀

Also, Magnification = -v/u

   ∴ H₁/H₀ = -v/u

⇒ H₁/4 = - (-37.5)/25

⇒ H₁ = 37.5 × 4/25

⇒ H₁ = 150/25

⇒ H₁ = 6 cm.

Since, the Size of the Image is greater than the size of the Object, thus the image is Magnified.

Hence, Characteristics of the Images are---

Real, Inverted and Magnified.

Hope it helps.