Your equations for $f$ and $g$ are correct. If these two lines intersect, then $f(x) = g(x)$, or $$ \tan \alpha (x + x_p) + y_p = \cot \beta (x + x_p) + y_p $$ $$ (\tan \alpha - \cot \beta ) (x + x_p) = 0. $$ If this equation is to hold regardless of the choice of $\alpha$ and $\beta$, which are independent, $x + x_p$ must be zero. Otherwise, changing $\alpha$ or $\beta$ will cause the left hand side to change, but not the right hand side. Now, with $x = -x_p$, you will find $$ y = f(-x_p) = g(-x_p) = -y_p. $$ Therefore the image you are after is located at $(-x_p, -y_p)$. However, there is a fundamental problem with your approach. At point $P'''$, there are really two overlapping images: the image through mirror 1 of the image through mirror 2, and the image through mirror 2 of the image through mirror 1. When the mirrors are perpendicular, these two images are located at the same point: $(-x_p, -y_p)$. This is why setting $f(x) = g(x)$ worked in your approach: these lines go through the same point. However, when the mirrors are not perpendicular, these two images will appear at different points, which complicates things. What you should realize is that $y = f(x)$ defines a family of lines passing through the image through the horizontal mirror of the image through the vertical mirror, so $f(x)$ by itself is sufficient to determine the image coordinates. For clarity, let $f_\alpha(x)$ be equal to $f(x)$, with the subscript indicating the parameter $\alpha$. The location of the image doesn't depend on $\alpha$, so for two different angles $\alpha$ and $\alpha '$, we have $$f_\alpha (x) = f_{\alpha '} (x) $$ $$ \tan \alpha (x + x_p) - y_p = \tan \alpha ' (x + x_p) - y_p $$ $$ (\tan \alpha - \tan \alpha ')(x + x_p) = 0. $$ For the same reasons as before, this equation requires $x = -x_p$, and then $y = f(-x_p) = -y_p$. This solution can be repeated for $g(x)$, which will yield the same coordinates, meaning the two images are co-located. Finally, as for your question about what happens if $f(x)$ and $g(x)$ have the same slope: this only happens when you shoot the beam right at the intersection of the two mirrors. In this case, the lines defined by $f(x)$ and $g(x)$ overlap. They still go through the location of the images, and there are no problems. Open in App Suggest Corrections |
What will happen to the image if two plane mirrors are held perpendicular to each other also comment on the no images formed?
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