Answer
Verified
110.1k+ views
Hint If we want to magnify an image of an object, we use a spherical mirror called a concave mirror. Next, we need to recall the conventions which we use for making ray diagrams involving concave mirrors. The object height is always taken positive if the object is in the positive y-direction.
Complete step by step answer
Spherical mirrors: it is a part of a spherical surface mirror. It is of two types- one is concave and another one is convex. The bulged out part is known as convex mirror whereas it is thinner from the center and thicker at the edges then it is known as concave mirror.
Magnification is defined as the ratio of image height to an object height. It is denoted by ‘m’.
$m = \dfrac{{{h_i}}}{{{h_o}}}$ Where ${h_{i\,}}$ -image height; ${h_o}$ - object height; m- magnification.
It is used to increase the height of the object.
Real image: when an image is formed on the screen.
Virtual image: when an image appears to form on the screen.
Image gets shorter and longer depending upon mirrors, but the object is always placed above the principal axis.
So, it will always be positive. Therefore, option B is correct.
Note
If object height is taken as negative then it should be placed below the principal axis, but that is against the well-established convention, so it is unwise to think in that direction. Hence option A is considered wrong.
Complete step by step answer
Spherical mirrors: it is a part of a spherical surface mirror. It is of two types- one is concave and another one is convex. The bulged out part is known as convex mirror whereas it is thinner from the center and thicker at the edges then it is known as concave mirror.
Magnification is defined as the ratio of image height to an object height. It is denoted by ‘m’.
$m = \dfrac{{{h_i}}}{{{h_o}}}$ Where ${h_{i\,}}$ -image height; ${h_o}$ - object height; m- magnification.
It is used to increase the height of the object.
Real image: when an image is formed on the screen.
Virtual image: when an image appears to form on the screen.
Image gets shorter and longer depending upon mirrors, but the object is always placed above the principal axis.
So, it will always be positive. Therefore, option B is correct.
Note
If object height is taken as negative then it should be placed below the principal axis, but that is against the well-established convention, so it is unwise to think in that direction. Hence option A is considered wrong.
Recently Updated Pages
Write an article on the need and importance of sports class 10 english JEE_Main
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
If x2 hx 21 0x2 3hx + 35 0h 0 has a common root then class 10 maths JEE_Main
The radius of a sector is 12 cm and the angle is 120circ class 10 maths JEE_Main
For what value of x function fleft x right x4 4x3 + class 10 maths JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
In Searles apparatus when the experimental wire is class 11 physics JEE_Main
The energy stored is a condenser is in the form of class 12 physics JEE_Main
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main