What will be ratio `(D"/"f)` in microscope where, D is the diameter of the aperture and .f. is the focal length of the objective lens?

Options

A. `tan beta`
B. `tan (beta)/(2)`
C. `2tan beta`
D. `tan (beta)/(6)`

Answer

To find the ratio \( \frac{D}{f} \) in a microscope, where \( D \) is the diameter of the aperture and \( f \) is the focal length of the objective lens, we can follow these steps:

### Step 1: Understand the components of a microscope
A microscope consists of two main lenses: the objective lens and the eyepiece. The objective lens has a focal length \( f \) and is responsible for forming a magnified image of the object being observed. The diameter of the aperture \( D \) affects the amount of light that can enter the microscope and thus impacts the resolution and brightness of the image.

### Step 2: Relate the diameter of the aperture to the focal length
In optical systems, the diameter of the aperture \( D \) is often related to the focal length \( f \) through the numerical aperture (NA) of the lens. The numerical aperture is defined as:
\[ NA = \frac{n \cdot \sin(\theta)}{f} \]
where \( n \) is the refractive index of the medium and \( \theta \) is the half-angle of the maximum cone of light that can enter the lens.

### Step 3: Express the ratio \( \frac{D}{f} \)
For a microscope, the diameter of the aperture can be approximated as:
\[ D \approx 2f \cdot \sin(\theta) \]
Thus, we can express the ratio \( \frac{D}{f} \) as:
\[ \frac{D}{f} \approx 2 \sin(\theta) \]

### Step 4: Conclusion
The ratio \( \frac{D}{f} \) is dependent on the angle \( \theta \) which defines the light cone entering the objective lens. Therefore, we can conclude that:
\[ \frac{D}{f} \approx 2 \sin(\theta) \]

### Final Answer
The ratio \( \frac{D}{f} \) in a microscope is approximately \( 2 \sin(\theta) \).

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