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Magnification In Lenses - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Magnification in Lenses is considered one of the most asked concept.
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13 Questions around this concept.
Solve by difficulty
A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in a normal setting, the angle formed by the image of the tower is , then
close to :
An object is placed at the focus of concave lens having focal length $f$. What is the magnification and distance of the image from the optical centre of the lens?
A convex lens produces an image of a real object on a screen with a magnification of . When the lens is moved
away from the object, the magnification of the image on the screen is
. The focal length of the lens is:
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An object is placed at a distance 'd' from convex lens of focal length 'f' the magnificatory of the image will be
A 16 cm long image of an object is formed by a convex lens on a screen placed normal to its principal axis. On moving the lens towards the screen, without changing the positions of the object and the screen, a 9 cm long image is formed again on the screen. The size of the object is:
A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in
A. a larger angle to be subtended by the object at the eye and hence viewed in greater detail.
B. the formation of a virtual erect image.
C. increase in the field of view.
D. infinite magnification at the near point.
Concepts Covered - 1
Magnification in Lenses
Magnification in Lenses-
Magnification produced by a lens is defined as the ratio of the size of the image to that of the object.
For the derivation let an object is placed whose height is h and the image height is h'. Let us consider an object is placed on the principal axis with its height h perpendicular to the principal axis as shown in the figure. The first ray passing through the optic centre will go undeviated. The second ray parallel to the principal axis must pass through the focus F' The image is formed where both the light rays intersect. By this, we get an image of height h'. So mathematically, magnification can be written as -
$\mathbf{m}=\frac{\mathbf{h}^{\prime}}{\mathbf{h}}$
Like the mirror equation, we can derive the formula of magnification in terms of the position of the object and image. For this, consider the figure given below -
As the triangle, ABO and $\mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{O}$ is similar so we can write that -
$
\frac{A^{\prime} B^{\prime}}{A B}=\frac{O B^{\prime}}{O B}
$
So,
$
\begin{aligned}
& \frac{h^{\prime}}{h}=\frac{v}{u} \\
& m=\frac{h^{\prime}}{h}=\frac{v}{u}
\end{aligned}
$
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