Homework 5 hints
I'll start off with number 1 and keep adding to this throughout the evening.
1. The apparent depth is where the light appears to have come from under the water (according to the observer) if refraction were not present. Extrapolate the light back from the observer, ignore refraction, and when the extrapolated ray reaches the same horizontal position as the actual object, you're at the apparent depth. Draw a little picture, it will help.
a. The angle of incidence can be related to the angle the (real) ray in the water makes with the surface via Snell's law. Once you have this angle, since you know the real depth you can find the lateral (horizontal) position of the object with some trigonometry. Now worry about the extrapolated ray under the water. You know its angle with respect to the surface is the same as the ray above the water by construction. Using the lateral position of the object and this angle, you can find the apparent depth.
b. You can generalize your result from a if you are careful about setting the angle to zero, or you can realize this is the same as a spherical surface with an infinite radius. For a spherical surface you know $n_1/p+n_2/q=(n_2-n_1)/R$. If $R$ tends to infinity, this gives $n_1/p+n_2/q$. The apparent depth is $|q|$, the real depth $p$, and that's that.
2. You are given $p+q=L$, and the lens equation tells you $1/p + 1/q = 1/f$. What will determine the minimum value for L? If we want to focus an image, we need to have a real image, not a virtual one. For a convex lens, this will happen only when $p>f$. We also know that $q$ must be positive for the image to be real. Combining these two facts, we really only need to make sure that $q$ is real and positive in order to have a real image formed. If $q$ is negative, the image is virtual. If $q$ is imaginary, so is the image ...
What you need to do, then, is solve one equation for $q$ and put it in the other. The resulting expression for $q$ will be quadratic, the condition that the roots are real is that the discriminant is positive. Start by noting that $1/p + 1/q = 1/f$ can be rewritten as $1/f = (p+q)/pq$. Knowing $p+q=L$, change the numerator. Knowing that $p=q-L$ (since $p+q=L$), change the denominator. Solve for $q$ in terms of $p$ and $L$. Do that stuff I just mentioned above. A "4" should be involved.
3. The previous problem should have given you two possible solutions for $q$ that give the minimum $L$. Those solutions should be solutions to the quadratic equation, and look something like: $q\in\{q_1,q_2\}=\{x+y, x-y\}$ where $x$ and $y$ are quantities involving $L$ and $f$ (and likely a "4"). The "$y$" quantity is likely to look like $\sqrt{\ldots}$. Those are your two possible $q$ positions, which imply two possible corresponding $p$ positions $p\in\{p_1,p_2\}=\{L-q_1,L-q_2\}$.
You know the magnification factor in each case in terms of the q's and p's. In the first case, $M_1=-q_1/p_1=a/h$, and in the second $M_2=-q_2/p_2=b/h$. Solve both of those for $h$, and multiply them to get an equation for $h^2$. In the resulting equation, you will have to show that $p_1p_2/q_1q_2=1$. Use your equations for $q$ above and multiply your two solutions together, cancellations will occur and it will end up simply. Do the same for the $p$'s, noting that $p_1=L-q_1$ and $p_2=L-q_2$. If you are careful in your algebra, the desired result will follow.
4. a. There is only one sort of spherical mirror that always gives virtual images. For any spherical mirror, the focal point is at a distance $R/2$. However: if the (virtual) image forms on the opposite side of the mirror as the image, what does that imply about the sign of the focal length? Remember the sign conventions!
b. You know $M=-q/p$, or $q=-Mp$. You also know $1/p+1/q=1/f$. Taking careful notice of the sign of $f$, you can plug in $q=-Mp$ and $|f|=R/2$ and solve this thing for $p$ in terms of $M$ and $R$, both of which are known. Your value for $p$ should come out to be positive.
1. The apparent depth is where the light appears to have come from under the water (according to the observer) if refraction were not present. Extrapolate the light back from the observer, ignore refraction, and when the extrapolated ray reaches the same horizontal position as the actual object, you're at the apparent depth. Draw a little picture, it will help.
a. The angle of incidence can be related to the angle the (real) ray in the water makes with the surface via Snell's law. Once you have this angle, since you know the real depth you can find the lateral (horizontal) position of the object with some trigonometry. Now worry about the extrapolated ray under the water. You know its angle with respect to the surface is the same as the ray above the water by construction. Using the lateral position of the object and this angle, you can find the apparent depth.
b. You can generalize your result from a if you are careful about setting the angle to zero, or you can realize this is the same as a spherical surface with an infinite radius. For a spherical surface you know $n_1/p+n_2/q=(n_2-n_1)/R$. If $R$ tends to infinity, this gives $n_1/p+n_2/q$. The apparent depth is $|q|$, the real depth $p$, and that's that.
2. You are given $p+q=L$, and the lens equation tells you $1/p + 1/q = 1/f$. What will determine the minimum value for L? If we want to focus an image, we need to have a real image, not a virtual one. For a convex lens, this will happen only when $p>f$. We also know that $q$ must be positive for the image to be real. Combining these two facts, we really only need to make sure that $q$ is real and positive in order to have a real image formed. If $q$ is negative, the image is virtual. If $q$ is imaginary, so is the image ...
What you need to do, then, is solve one equation for $q$ and put it in the other. The resulting expression for $q$ will be quadratic, the condition that the roots are real is that the discriminant is positive. Start by noting that $1/p + 1/q = 1/f$ can be rewritten as $1/f = (p+q)/pq$. Knowing $p+q=L$, change the numerator. Knowing that $p=q-L$ (since $p+q=L$), change the denominator. Solve for $q$ in terms of $p$ and $L$. Do that stuff I just mentioned above. A "4" should be involved.
3. The previous problem should have given you two possible solutions for $q$ that give the minimum $L$. Those solutions should be solutions to the quadratic equation, and look something like: $q\in\{q_1,q_2\}=\{x+y, x-y\}$ where $x$ and $y$ are quantities involving $L$ and $f$ (and likely a "4"). The "$y$" quantity is likely to look like $\sqrt{\ldots}$. Those are your two possible $q$ positions, which imply two possible corresponding $p$ positions $p\in\{p_1,p_2\}=\{L-q_1,L-q_2\}$.
You know the magnification factor in each case in terms of the q's and p's. In the first case, $M_1=-q_1/p_1=a/h$, and in the second $M_2=-q_2/p_2=b/h$. Solve both of those for $h$, and multiply them to get an equation for $h^2$. In the resulting equation, you will have to show that $p_1p_2/q_1q_2=1$. Use your equations for $q$ above and multiply your two solutions together, cancellations will occur and it will end up simply. Do the same for the $p$'s, noting that $p_1=L-q_1$ and $p_2=L-q_2$. If you are careful in your algebra, the desired result will follow.
4. a. There is only one sort of spherical mirror that always gives virtual images. For any spherical mirror, the focal point is at a distance $R/2$. However: if the (virtual) image forms on the opposite side of the mirror as the image, what does that imply about the sign of the focal length? Remember the sign conventions!
b. You know $M=-q/p$, or $q=-Mp$. You also know $1/p+1/q=1/f$. Taking careful notice of the sign of $f$, you can plug in $q=-Mp$ and $|f|=R/2$ and solve this thing for $p$ in terms of $M$ and $R$, both of which are known. Your value for $p$ should come out to be positive.
2 comments:
Can you please post the solutions to HW5 after midnight
I will post them tonight, sure.
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