What angle is subtended when using this 270-diopter objective?

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The discussion centers on calculating the angle subtended by an image when using a 270-diopter objective lens, given that a 140-diopter lens subtends an angle of 3.7 x 10^-3 rad. The initial attempt to solve the problem involved using the tangent function, but it was noted that the small angle approximation could simplify the process. By recognizing that the angle will increase proportionally to the ratio of the objective strengths, the angle for the 270-diopter lens can be calculated by multiplying the original angle by the ratio of the two diopter values. This approach clarifies the relationship between the objective strength and the subtended angle. The correct angle can be derived easily by applying this proportionality principle.
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Homework Statement


An anatomist is viewing heart muscle cells with a microscope that has two selectable objectives with refracting powers of 140 and 270 diopters. When she uses the 140-diopter objective, the image of a cell subtends an angle of 3.7 x 10-3 rad with the eye. What angle is subtended when she uses the 270-diopter objective?


Homework Equations


tan alpha=h/f


The Attempt at a Solution


h=tan(3.7x10^-3)(140) = 0.518
tan beta = 0.518/270
beta = 0.0019185 rad


This is incorrect... I am not sure if I'm even going about this problem the right way. Thanks for any help you can give!
 
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Note the small angle involved (3.7 x 10^-3 rad), so the small angle approximation is valid
θ ≈ sinθ ≈ tanθ​
The image will simply be magnified in proportion to the power of the objective.
 
I don't really understand what this means.. how are you saying I would go about solving this with that in mind?
 
The image originally subtended an angle of 3.7 x 10^-3 rad.

Using a stronger objective, we expect the image to appear larger.

Since the objective strength has been increased by a ratio of (270/140), the 3.7 x 10^-3 rad angle will get multiplied by that same factor.
 
Ohh, I was making it harder than it should be. Thank you very much!
 
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Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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