An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.
An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.
Given, height of object = 5cm
Position of object, u = - 25cm
Focal length of the lens, f = 10 cm
Hence, position of image, v =?
We know that,
1/v - 1/u = 1/f
1/v + 1/25 = 1/10
So, 1/v = 1/10 - 1/25
S0, 1/v = (5 - 2)/50
That is, 1/v = 3/50
So, v= 50/3 = 16.66 cm
Thus, distance of image is 16.66 cm on the opposite side of lens.
Now, magnification = v/u
That is, m = 16.66/-25 = -0.66
Also, m= height of image/height of object
Or, -0.66 = height of image / 5 cm
Therefore, height of image = -3.3 cm
The negative sign of height of image shows that an inverted image is formed.
Thus, position of image = At 16.66 cm on opposite side of lens
Size of image = - 3.3 cm at the opposite side of lens
Nature of image – Real and inverted...
Given, height of object = 5cm
Position of object, u = - 25cm
Focal length of the lens, f = 10 cm
Hence, position of image, v =?
We know that,
1/v - 1/u = 1/f
1/v + 1/25 = 1/10
So, 1/v = 1/10 - 1/25
S0, 1/v = (5 - 2)/50
That is, 1/v = 3/50
So, v= 50/3 = 16.66 cm
Thus, distance of image is 16.66 cm on the opposite side of lens.
Now, magnification = v/u
That is, m = 16.66/-25 = -0.66
Also, m= height of image/height of object
Or, -0.66 = height of image / 5 cm
Therefore, height of image = -3.3 cm
The negative sign of height of image shows that an inverted image is formed.
Thus, position of image = At 16.66 cm on opposite side of lens
Size of image = - 3.3 cm at the opposite side of lens
Nature of image – Real and inverted...
