Question

(d) m = 2/3 for the hand and m = 1/3 for the face

4. Parmigianino's Self Portrait The Italian artist Parmigianino (Francesco Mazzola) displayed a 16th century fascination with mirrors with his Self Portrait in a Convex Mirror (c. 1524). (a) Which of Francesco's hands held the paint brush (left or right)? Explain. (Assume that the brush is in the hand which is not visible in the painting.) (b) The image is distorted, in that the hand appears relatively large in comparison to the face Explain this effect by drawing two high-quality ray diagrams, in the first case with dof/2 and in the second case with do--2f (c) Use your ray diagrams to estimate the magnification which one would obtain in each of the two cases considered in part (b). (Explain your estimate.) (d) Check your estimated magnifications from (c) by calculating the exact magnifications algebraically, for each of the two cases.

Answer 1

A convex mirror is a spherical mirror where the outer side of the surface is reflective. It is quite similar to plane mirror in its working, in that the image is always on the other side of the mirror while the size of the image never exceeds or even reaches the size of the object. A plane mirror has the image have the size as the object.

Another similarity is the lateral inversion of the object (i.e. Left becomes right and right becomes left).

(A) So, in this case, Francesco's painting arm which is holding the brush and not visible to us is on the right in the image. That means he painted with his left hand in real llife. This lateral inversion is a property of all mirrors.

(B) Any image created by the convex mirror can be encapsulated into a few rules,

1. Every image created by a convex mirror is virtual (i.e. behind the mirror) and hence cannot be projected on a screen.
2. The image distance increases with object distance but it never exceeds the focal length of the mirror, and when light from infinity hits a convex mirror, the object is a point on the virtual focus.
3. The image height decreases with the increase in object distance from the mirror surface. It reaches half the height of the object when the object is placed at the focal length.

That's it. Using this information, we can say that the hand, since it was closer to the mirror appears bigger than the face of Francesco. Were is on the focus, the face would be exactly half the height of his real face. So, we take two cases, on where the object (our hand) is at a distance d₀=-f/2 and where our object (the face) is at a distance d₀=-2f.

We will start with our radius of curvature is twice the focal length for our mirror. Consider a mirror with focal length = -4 cm, where we have the object at half the focal length at 2 cm with height 2 cm. The image will have to be between the origin and the object distance. By drawing all the rays (the normal ray, the parallel ray and the diverging ray), we estimate the image to be at around d_i ~ -1.5 cm.This is because the height has to be greater than 1 cm since our rules say that,

The resultant ray diagram is,

Now, for the second case d₀ = -2f = 8 cm. For this we can estimate that image formed will be farther than the previous image but still be less than the focal length (according to our rules). According to our esitmate is comes out to be d_i ~ -3 cm. The height of the image will also be less than half of the object height because of equation (1). The resultant estimated ray diagram is,

(C) So it is quite clear that,

$h_i(face)<h_i(hands)$
$d_i(face)>d_i(hands)$

And hence magnification is given by,

$m(hands) = -\frac{d_i}{d_0} = \frac{1.5}{2} = 0.75$

$m(face) = -\frac{d_i}{d_0} = \frac{3}{8} = 0.375$

This is why the hands appear bigger.

(D) A much better way of figuring out image distance is by using the mirror formula, which states that,

$\frac{1}{d_0}+\frac{1}{d_i} = \frac{1}{f}$

So, our focal length was f = - 4 cm (the negative sign is due to the sign convention because the focus is on the inside of a convex mirror). And our object distance was, d₀ = -(-4)/2 = 2 cm, then the image distance will be,

$\frac{1}{2}+\frac{1}{d_i} = \frac{1}{-4}$

${\color{Red}d_i = -1.33\ cm }$

which is close to our estimate of 1.5 cm on the inside of the mirror.

Similarly for the second case, we can write,

$\frac{1}{8}+\frac{1}{d_i} = \frac{1}{-4}$

${\color{Red} d_i = -2.67\ cm}$

Algebraic values for magnification are,

$m(hands) = -\frac{d_i}{d_0} = \frac{1.33}{2} = 0.665$

$m(face) = -\frac{d_i}{d_0} = \frac{2.67}{8} = 0.33$

So we were quite close with our estimation although not as exact as the mirror formula. Anyway, we can see that as the distance increases magnification decreases and image distance approaches the focal length .