Question:

The magnification of a concave lens is always

Updated On: Apr 17, 2025

equal to one
less than one
greater than one
none of the above

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The Correct Option is B

Solution and Explanation

To solve the problem, we need to understand the behavior of a concave lens.

1. Magnification of a Concave Lens:
A concave lens always forms: - A virtual image, - An upright image, - A diminished image (smaller than the object).

2. Conclusion:
The magnification of a concave lens is always less than one because the image formed is smaller than the object.

Final Answer:
The magnification of a concave lens is always $ {\text{less than one}} $.

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Top Questions on Reflection Of Light By Spherical Mirrors

A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence $\theta_0$. The ray partially refracts into the sphere with angle of refraction $\phi$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $a$. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-I		List-II
P	If $n = 2$ and $\alpha = 180°$, then all the possible values of $\theta_0$ will be	I	$30\degree$ or $0\degree$
Q	If $n = √3$ and $\alpha= 180°$, then all the possible values of $\theta_0$ will be	II	$60\degree$ or $0\degree$
R	If $n = √3$ and $\alpha= 180°$, then all the possible values of $\phi_0$ will be	III	$45\degree$ or $ 0\degree$
S	If $n = \sqrt2$ and $\theta_0 = 45°$, then all the possible values of $\alpha$ will be	IV	$150\degree$
			\[0\degree\]

JEE Advanced - 2024
Physics
Reflection Of Light By Spherical Mirrors

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List-I		List-II
P	If \(n = 2\) and \(\alpha = 180°\), then all the possible values of \(\theta_0\) will be	I	\(30\degree\) or \(0\degree\)
Q	If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\theta_0\) will be	II	\(60\degree\) or \(0\degree\)
R	If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\phi_0\) will be	III	\(45\degree\) or \( 0\degree\)
S	If \(n = \sqrt2\) and \(\theta_0 = 45°\), then all the possible values of \(\alpha\) will be	IV	\(150\degree\)
			\[0\degree\]