A square wire of side 3 cm is placed 25 cm away from a concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire? The centre of the wire is on the axis of the with it's two sides normal to the axis.
Answers
Given :-
- Side of the square wire = 3 cm
- object distance = 25 cm
- Focal length of the concave mirror = 10 cm.
- The centre of the wire is on the axis of the with it's two sides normal to the axis.
To Find :-
- The area enclosed by the image of the wire
Solution :-
According to the Question,
Type of Mirror = Concave Mirror
Focal Length, f = -10cm
Object distance, u = -25 cm
Height of object, ho = 3cm
Using mirror formula,
1/f = 1/u + 1/v
↠ 1/-10 = 1/-25 + 1/v
↠ -1/10 = -1/25 + 1/v
↠ -1/10 + 1/25 = 1/v
↠ -5+2/50 = 1/v
↠ -3/50 = 1/v
↠ v = -50/3 cm
Now,
m = hi/ho = -v/u
↠ hi/3 = -(-50/3)/-25
↠ hi/3 = -(50/3) /25
↠ hi= 3 × -(50/3)/25
↠hi = -50/25
↠ hi = -2cm
↠ Area = (-2m) × (-2m)
↠ Area = 4m²
Hence, the area enclosed by the image of the square wire is 4m².
- Given:-
A square wire of side 3 cm is placed 25 cm away from a concave mirror of focal length 10 cm. The centre of the wire is on the axis of the with it's two sides normal to the axis.
To Find :-
What is the area enclosed by the image of the wire?
Solution:-
According to the Question
Firstly we will calculate the image position of the object (wire) .
Type of Mirror = Concave Mirror
As we know that ,
Focal Length ,f = -10cm
Object distance ,u = -25 cm
Height of object ,ho = 3cm
by using mirror formula
1/f = 1/u + 1/v
substitute the given value we get
↠ 1/-10 = 1/-25 + 1/v
↠ -1/10 = -1/25 + 1/v
↠ -1/10 + 1/25 = 1/v
↠ -5+2/50 = 1/v
↠ -3/50 = 1/v
↠ v = -50/3 cm
Now, calculating the height of the image by using magnification formula
m = hi/ho = -v/u
putting the value we get
↠ hi/3 = -(-50/3)/-25
↠ hi/3 = -(50/3) /25
↠ hi= 3 × -(50/3)/25
↠hi = -50/25
↠ hi = -2cm
Now, calculating the area enclosed by the image of the square wire
↠ Area = (-2m) × (-2m)
↠ Area = 4m²
Hence, the area enclosed by the image of the square wire is 4m².