A lady cannot see objects closer than `40cm` from the left eye and closer than `100cm` from the right eye. While on a mountaining trip, she is lose from her team. She tries to make an astronomical trip, from her reading glasses to look from her teammates. (a) Which glass should she use as the eyepiece? (b) What magnification can she get with relaxed eye?

Answer

To solve the problem step by step, we need to determine the appropriate eyepiece for the lady's astronomical telescope and calculate the magnification she can achieve with her relaxed eye.

### Step 1: Identify the Near Points
The problem states:
- The near point for the left eye is \(40 \, \text{cm}\).
- The near point for the right eye is \(100 \, \text{cm}\).

### Step 2: Calculate the Focal Length for Each Eye
We will use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \(f\) is the focal length of the lens,
- \(v\) is the image distance (which we can take as \(-25 \, \text{cm}\) for a relaxed eye),
- \(u\) is the object distance (the near point).

#### For the Left Eye:
Using the near point of \(40 \, \text{cm}\):
\[ \frac{1}{f_1} = \frac{1}{-25} - \frac{1}{40} \]
Calculating this:
\[ \frac{1}{f_1} = -\frac{1}{25} + \frac{1}{40} \]
Finding a common denominator (100):
\[ \frac{1}{f_1} = -\frac{4}{100} + \frac{2.5}{100} = -\frac{1.5}{100} = -\frac{3}{200} \]
Thus,
\[ f_1 = -\frac{200}{3} \, \text{cm} \]

#### For the Right Eye:
Using the near point of \(100 \, \text{cm}\):
\[ \frac{1}{f_2} = \frac{1}{-25} - \frac{1}{100} \]
Calculating this:
\[ \frac{1}{f_2} = -\frac{1}{25} + \frac{1}{100} \]
Finding a common denominator (100):
\[ \frac{1}{f_2} = -\frac{4}{100} + \frac{1}{100} = -\frac{3}{100} \]
Thus,
\[ f_2 = -\frac{100}{3} \, \text{cm} \]

### Step 3: Determine Which Lens to Use as the Eyepiece
In an astronomical telescope, the objective lens has a longer focal length than the eyepiece. Here:
- \(f_1 = -\frac{200}{3} \, \text{cm}\) (left eye)
- \(f_2 = -\frac{100}{3} \, \text{cm}\) (right eye)

Since \(f_1 < f_2\), we use:
- Left eye lens as the **objective** (focal length \(f_1\))
- Right eye lens as the **eyepiece** (focal length \(f_2\))

### Step 4: Calculate the Magnification
The magnification \(M\) of an astronomical telescope is given by:
\[ M = -\frac{f_1}{f_2} \]
Substituting the values:
\[ M = -\left(-\frac{200}{3}\right) / \left(-\frac{100}{3}\right) = \frac{200}{100} = 2 \]

### Final Answers
(a) The eyepiece should be the **right eye lens**.
(b) The magnification she can get with a relaxed eye is **2**.