Question

Under which of the following conditions a concave mirror can form a real image larger than the actual object?
A. When the source is at a distance equal to the radius of curvature of the concave mirror.
B. when the source is at a distance less than the focal length of the concave mirror.
C. when the source is placed at a distance between the focus and centre of curvature of the concave mirror.
D. when the source is placed at a distance greater than the radius of curvature of the concave mirror.

Answer 1

Hint: In a concave mirror, the object is real, that is we need to imagine the magnification accordingly and proceed with the problem and also notice that the question says that the real image is larger than the actual object.

Complete answer:
In a concave mirror, the image is real so the magnification is negative and we know that the magnification when the image is larger than the object is $m=-\dfrac{v}{u}$ and also, the mirror formula, $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$. And so, the equation becomes $v=\dfrac{uf}{u-f}$.

So, from both the equations, we can see that if we put the value of $v$ in the magnification equation, the equation becomes $m=\dfrac{f}{f-u}$. Now we use this formula and check which of the options fits in the equation.

In the first case, it is given that $u=-2f$ so, if we solve, the equation gives us that magnification is $-1$ that is the image is of the same size so it is not the answer as $\left| m \right|>1$ for the answer. In the second case, we can assume that $u=-\dfrac{f}{2}$ and accordingly, $m$ becomes +2 that is the image is virtual so the option is rejected. In the 3rd case take $u=-\dfrac{3f}{2}$ and put it in the magnification equation, $m$ comes out to be $-2$. That is the option is correct.

So the answer is option C.

Note: In option number D, we can take the object distance, $u$ as $3f$ and if we put that in the equation, we get $m$ as $-\dfrac{1}{2}$. That is $\left| m \right|>1$ condition is not satisfied and so the answer is option C.