What is the approximate magnification of a compound microscope with objective and eyepiece focal lengths of 0.3 cm and 3.6 cm, respectively, and a separation between lenses of 20 cm?
The answer is 460.
I tried finding the magnification for the eyepiece which was m=25/f=25/3.6=6.94.
For the objective lens I tried to solve for q by 1/20 + 1/q = 1/.3 and solved for magnification of objective lens to be m=-q/p.. but i didn't get the correct answer.. is the given value of 20cm=p? i think that's where i went wrong.
Assistance needed.
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A resonable approximation for magnification of a microscope is
M= L/f0 * 25/fe
See:http://hyperphysics.phy-astr.gsu.edu/hbasees/geoopt/micros.html
With that approximation, I get
M= 20/.3*25/3.6=463 and it has to be rounded to two sig figures.
Within a compound microscope, the objective and eyepiece have focal lengths of 0.9 cm and 2.5cm respectively. The distance between the eyepiece and objective is 23.2 cm. A real image formed by the objective is 17 cm from the objective. Where is the object located?
To find the magnification of a compound microscope, you need to consider the magnification of the objective lens and the eyepiece lens separately, and then multiply them together to get the total magnification.
Let's go through the calculations step by step:
1. Determine the magnification of the eyepiece lens (m_e):
Using the formula m_e = 25/f_e, where f_e is the focal length of the eyepiece lens, we have:
m_e = 25/3.6 = 6.94 (rounded to two decimal places).
2. Determine the distance between the objective and eyepiece lenses (p):
The given information tells us that the separation between lenses is 20 cm, so p = 20 cm.
3. Determine the magnification of the objective lens (m_o):
We can use the lens formula, 1/f_o = 1/p + 1/q, where f_o is the focal length of the objective lens, and q is the distance between the objective lens and the object being observed, which in this case is undefined.
Rearranging the lens formula, we have: q = fp / (f_o - p), where p = 20 cm and f_o = 0.3 cm.
Substituting the values, we get: q = (0.3 cm * 20 cm) / (0.3 cm - 20 cm) ≈ -0.59 cm.
Note that the negative sign indicates that the real object is located inside the focal point of the objective lens. However, since the question asks for the magnification, the sign will not be relevant in this case.
The magnification of the objective lens is given by m_o = -q/p ≈ -0.59 cm / 20 cm ≈ -0.0295 (rounded to four decimal places).
4. Calculate the total magnification (m_total):
To obtain the total magnification, we multiply the magnification of the objective lens by the magnification of the eyepiece lens: m_total = m_e * m_o ≈ 6.94 * -0.0295 ≈ -0.20413 (rounded to five decimal places).
The correct answer to the question should be the absolute value of the total magnification, as magnification is a positive quantity. Therefore, the approximate magnification of the compound microscope is 0.20413, which can be rounded to 0.20 or 200 when expressed as a percentage.
It is essential to check your calculations and consider the sign conventions in lens formula applications when calculating the magnification.