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प्रश्न
An object 60 cm from a lens gives a virtual image at a distance of 20 cm in front of the lens. What is the focal length of the lens? Is the lens converging or diverging? Give reasons for your answer.
उत्तर
Object distance u = - 60 cm (As the object is on the left side of the lens)
Image distance v = - 20 cm (As the image is virtual)
Magnification `m=v/u=(-20)/-60=1/3`
Because the image is diminished and virtual, the lens is concave and diverging in nature.
Using the lens formula:
`1/f=1/v-1/u`
`1/f=1/-20-1/-60`
`1/f=1/-20+1/60`
`1/f=(-3+1)/60=-1/30`
∴ f = - 30 cm.
Therefore, the focal length of the lens is 30 cm.
The lens is diverging in nature because a negative focal length indicates a concave lens .
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संबंधित प्रश्न
Define the principal focus of a concave lens.
State whether concave lens has a real focus or a virtual focus.
List some things that concave lens and concave mirror have in common.
What type of image is always made by a concave lens?
Construct ray diagrams to illustrate the formation of a virtual image using a diverging lens.
(a) Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from:
(i) a diverging lens of focal length 40 cm.
(ii) a converging lens of focal length 40 cm.
(b) Draw labelled ray diagrams to show the formation of images in case (i) and (ii) above (The diagrams may not be according to scale).
What kind of lens is used to correct
short-sightedness
An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm.
(i) Use the lens formula to find the distance of the image from the lens.
(ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case.
(iii) Draw a ray diagram to justify your answer of the part (ii).
An object is placed in front of a lens between its optical centre and focus. The formed image is virtual, erect, and diminished. Name the lens used.
In a reading glass what is the position of the object with respect to the concave lens used?
