Question

An object 4cm in size is placed 25cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature ans size of the image.

Answer 1

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,

    $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
⇒  $\frac{1}{-15} = \frac{1}{-25} + \frac{1}{v}$
⇒  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.

Answer 2

Given,
size of object , H₀ = 4cm
object distance , u = -25cm
focal length of concave mirror , f = -15cm

Use formula ,
1/v + 1/u = 1/f
⇒1/v + 1/-25 = 1/-15
⇒1/v = 1/-15 + 1/25 = (-25 + 15)/375 = 1/-37.5 cm
⇒ v = - 37.5 cm ,

∴ image is formed object side , 37.5cm from pole of concave mirror.
Now, use formula,
Hi/H₀ = -u/v
Where Hi is the size of image .
Now, Hi/4cm = -(-37.5cm)/(-25cm)
Hi = -4 × 37.5/25 cm = -6cm
∴ size of image = 6cm

Hence,sreen should place 37.5 cm left side of mirror and size of image is 6cm .it is inverted and larger than object's size.