5. An object 5 cm in length is held 25cm away from a converging lens of focal length 10cm. Draw the ray diagram and find the position, size and nature of the image formed.
Answers
Answer:
Given that
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:
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