Question

Magnifying power of an astronomical telescope is 'M.P'. If the focal length of objective is doubled, then the magnifying power for normal adjustment will become
(A) $2 \mathrm{M.P}$
(B) $\dfrac{\text { M.P }}{2}$
(C) $3 \mathrm{M.P}$
(D) $\dfrac{3 \mathrm{M.P}}{2}$

Answer 1

Hint: We know that magnification is the ability to make small objects seem larger, such as making a microscopic organism visible. Resolution is the ability to distinguish two objects from each other. When the image formed is virtual and erect, magnification is positive. And, when the image formed is real and inverted, magnification is negative. The underlying principle of a microscope is that lenses refract light which allows for magnification. Refraction occurs when light travels through an area of space that has a changing index of refraction.

Complete step by step answer
We know that the magnifying power is defined as the ratio between the dimensions of the image and the object. The process of magnification can occur in lenses, telescopes, microscopes and even in slide projectors. Simple magnifying lenses are biconvex - these lenses are thicker at the center than at the edges. The magnifying power, or extent to which the object being viewed appears enlarged, and the field of view, or size of the object that can be viewed, are related by the geometry of the optical system.
The magnifying power or M.P of the reflecting telescope is $\dfrac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}} .$
 If the focal length of the eye-piece is halved, then its magnifying power $\mathrm{m}^{\prime}\left(\dfrac{\mathrm{f}_{0} \times 2}{\mathrm{f}_{\mathrm{e}}}\right)$ is 2 M.P.

So, the correct answer is option A.

Note: We know that virtual images are always located behind the mirror. Virtual images can be either upright or inverted. Virtual images can be magnified in size, reduced in size or the same size as the object. Virtual images can be formed by concave, convex and plane mirrors. Magnification is the ratio between the height of the object and the height of the image. In case of a virtual image, the height of the image, as well as the height of the object, carries a positive sign. Therefore, the magnification is positive in this case. A real image occurs where rays converge, whereas a virtual image occurs where rays only appear to converge. Real images can be produced by concave mirrors and converging lenses, only if the object is placed further away from the mirror/lens than the focal point and this real image is inverted.