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Question: An object \(5cm\) in length is held \(25cm\) from a converging lens of focal length \(10cm\) .Draw t...

An object 5cm5cm in length is held 25cm25cm from a converging lens of focal length 10cm10cm .Draw the ray diagram and find the position, size, nature of the image formed.

Explanation

Solution

Hint As the height, distance of the object and the focal length of the object is given, use the distance formula and find the distance of the image and then calculate the magnification of the lens and calculate the height of the image and discuss the nature of the image.

Complete Step by step solution
Here the height of the object is 5cm5cm and the distance of the object from the focal point is 25cm25cm, the focal length of the lens is 10cm10cm .
We know the image distance formula, 1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} [where vvis the position of the image, uuis the distance of the image from focal length, ff is the focal length. ]
Now, u=−25u = - 25 and f=10f = 10
Hence, 1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}
Now we will substitute the given values as,
⇒1v−1−25=110\Rightarrow \dfrac{1}{v} - \dfrac{1}{{ - 25}} = \dfrac{1}{{10}}
After simplification we will get,
⇒1v=350\Rightarrow \dfrac{1}{v} = \dfrac{3}{{50}}
⇒v=16.66\Rightarrow v = 16.66
Hence the distance of the image is 16.66cm16.66cm.
Now we know the magnification is the ratio of distance of image and distance of object.
Hence, m=vu=16.66−25=−0.66m = \dfrac{v}{u} = \dfrac{{16.66}}{{ - 25}} = - 0.66
Similarly, we know the magnification is the ratio between the height of the image and height of the object.
m=h1h2⇒−0.66=h15⇒h1=−3.3m = \dfrac{{{h_1}}}{{{h_2}}} \Rightarrow - 0.66 = \dfrac{{{h_1}}}{5} \Rightarrow {h_1} = - 3.3
∴h1=−3.3cm\therefore {h_1} = - 3.3cm
Hence the height of the object is 3.3cm3.3cm and the negative sign shows that the image is inverted.
Hence, we can say the image is formed at 16.66cm16.66cm on the opposite sides of the lens.
Now the nature of the image is real and inverted.

Now we will draw the ray diagram of the image:

Note
As we know that two rays are used to draw the ray diagram while following the laws of reflection, it should be kept in mind that the process of constructing a ray diagram is the same for all the positions of objects while the result of the ray diagram is different.