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प्रश्न
उत्तर
Given, magnification,
m = -1
Since,
\[m = \frac{v}{u} = - 1\]
\[ \Rightarrow v = - u\]
This means that object and image formed is at the same distance from the optical centre of the lens and are on the opposite sides of the lens. Since it is given that the distance between the object and its image is 60 cm
\[\Rightarrow u = - 30 cm\]
\[and v = 30 cm\]
Hence, the distance of the object from the optical centre of the lens is 30 cm.
Using lens formula,
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]
\[\frac{1}{f} = \frac{1}{30} - \frac{1}{- 30} = \frac{2}{30}\]
\[f = 15 cm\]
The given lens is a convex lens as its focal length is positive and the focal length of this lens is 15 cm.
If the object is now displaced 20 cm towards the optical centre of the lens, that means it is now placed at a distance of 10 cm from the lens. This shows that the object is now placed between the optical centre and the focus of the lens. Using lens formula,
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]
\[\frac{1}{15} = \frac{1}{v} - \frac{1}{- 10} \]
\[\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{- 1}{30}\]
\[v = - 30 cm\]
So, in this case the image is formed at a distance of 30 cm from the lens on the same side as the object.
The nature of the image formed is
1) virtual
2) erect
3) Enlarged
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संबंधित प्रश्न
A student has obtained a point image of a distant object using the given convex lens. To find the focal length of the lens he should measure the distance between the :
(A) lens and the object only
(B) lens and the screen only
(C) object and the image only
(D) lens and the object and also between the object and the image
A student wants to project the image of a candle flame on a screen 60 cm in front of a mirror by keeping the flame at a distance of 15 cm from its pole.
(a) Write the type of mirror he should use.
(b) Find the linear magnification of the image produced.
(c) What is the distance between the object and its image?
(d) Draw a ray diagram to show the image formation in this case.
In order to obtain a magnification of, −1.5 with a concave mirror of focal length 16 cm, the object will have to be placed at a distance
(a) between 6 cm and 16 cm
(b) between 32 cm and 16 cm
(c) between 48 cm and 32 cm
(d) beyond 64 cm
Explain what is meant by a virtual, magnified image.
A student wants to project the image of a candle flame on a screen 80 cm in front of a mirror by keeping the candle flame at a distance of 20 cm from its pole.
- Which type of mirror should the student use?
- Find the magnification of the image produced.
- Find the distance between the object and its image.
- Draw a ray diagram to show the image formation in this case and mark the distance between the object and its image.
To determine focal length of a concave mirror a student obtains the image of a well lit distant object on a screen. To determine the focal length of the given concave mirror he needs to measure the distance between:
(A) mirror and the object
(B) mirror and the screen
(C) screen and the object
(D) screen and the object and also mirror and the screen
Solve the following example.
An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?
An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger as compared to the object?
A lens of focal length 5 cm is being used by Debashree in the laboratory as a magnifying glass. Her least distance of distinct vision is 25 cm.
- What is the magnification obtained by using the glass?
- She keeps a book at a distance 10 cm from her eyes and tries to read. She is unable to read. What is the reason for this?
The magnification by a lens is -3. Name the lens and state how are u and v related?
