An object 5 cm in length is held 30 cm away from a concave mirror of focal length 10 cm. Draw the ray diagram and find the position and size of the image formed.
Answers
Answer:
The height of object = 5cm
Position of object, u = – 25cm
The focal length of the lens, f = 10 cm
We need to find
The position of the image, v =?
Size of the image
Nature of the image
Formula
We know that
1/v – 1/u = 1/f
Substituting the known values in the above equation we get,
1/v + 1/25 = 1/10
=> 1/v = 1/10 – 1/25
=> 1/v = (5 – 2)/50
Hence, 1/v = 3/50
So, v= 50/3 = 16.66 cm
Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.
Now, we know that
Magnification = v/u
Hence, m = 16.66/-25 = -0.66
Also, we know that
m= height of image/height of the object
Or, -0.66 = height of image / 5 cm
Hence, height of image = -3.3 cm
The negative sign of the height of the image depicts that an inverted image is formed.
So, the position of image = At 16.66 cm on the opposite side of the lens
Size of image = – 3.3 cm at the opposite side of the lens
Nature of image – Real and inverted
Answer:
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
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