Answer
a) See the figures below.
b) +2
Work Step by Step
a)
In a concave mirror, the observer can see a magnified virtual image in front of his face (not behind him), as you see in the first figure below.
But in a converging lens, the virtual magnified image is behind him so he can't see the image, as you see in the second figure below.
We can also get an upside real image by this lens in front of the observer but you need here a screen to see this image. It occurs when the object is far from the focal point, as you see in the third figure below.
Note that the object and the observer are at the same place since the observer is doing makeup using the mirror or the lens.
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b) We know that the magnification for both morros and lens is given by
$$m=\dfrac{-d_i}{d_o}$$
So, we need to find the image for both of them.
Let's start with the concave mirror;
$$\dfrac{1}{f_1}=\dfrac{1}{d_{01}}+\dfrac{1}{d_{i1}}$$
Solving for $d_{i1}$;
$$\dfrac{1}{f_1}-\dfrac{1}{d_{01}}=\dfrac{1}{d_{i1}}$$
$$d_{i1}=\left[\dfrac{1}{f_1}-\dfrac{1}{d_{01}}\right]^{-1}$$
Plugging the known
$$d_{i1}=\left[\dfrac{1}{f}-\dfrac{1}{\frac{1}{2}f}\right]^{-1}=-f\tag 1$$
By the same approach for the lens;
$$d_{i2}=\left[\dfrac{1}{f_2}-\dfrac{1}{d_{02}}\right]^{-1}$$
Plugging the known
$$d_{i2}=\left[\dfrac{1}{f}-\dfrac{1}{\frac{1}{2}f}\right]^{-1}=-f\tag 2$$
Since the object is at the same place for both of them and the image is also at the same, they have the same magnification factor.
$$m=\dfrac{-(-f)}{\frac{1}{2}f}=\bf 2$$
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