an object 4.0 CM in size is placed at 25.0 CM in front of a concave mirror of focal length 15.0 CM. at what distance from the mirror should a screen we placed in order to obtain a sharp image ? find the nature and size of the image?
Answers
Explanation:
The screen should be placed at a distance of 37.5 cm on the object side of the mirror, to obtain a sharp image of the object. The image formed is real, inverted (because h' is negative) and enlarged in size.
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,
⇒
⇒ 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.
an object 4.0 CM in size is placed at 25.0 CM in front of a concave mirror of focal length 15.0 CM. at what distance from the mirror should a screen we placed in order to obtain a sharp image ? find the nature and size of the image?
_____________________________
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.