A convexo-convex has a focal length of `f_1=10 cm.` One of the lens surfaces having a radius of curvature of `R=10 cm` is coated with silver. Construct the image of the object produced by the given optical system and determine the position of the image if the object is at a distance of `a=15 cm` from the lens. Refractive index of lens `=1.5.`

Explanation

To solve the problem step by step, we will use the lens formula and the information provided in the question.

### Step 1: Understand the Given Data
- Focal length of the lens, \( f_1 = 10 \, \text{cm} \)
- Radius of curvature of one surface, \( R = 10 \, \text{cm} \)
- Object distance, \( a = 15 \, \text{cm} \) (which we will take as \( u = -15 \, \text{cm} \) since the object is on the left side of the lens)
- Refractive index of the lens, \( n_2 = 1.5 \)
- Refractive index of air, \( n_1 = 1 \)

### Step 2: Use the Lens Maker's Formula
The lens maker's formula is given by:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a convexo-convex lens, we have:
- \( R_1 = +10 \, \text{cm} \) (convex surface)
- \( R_2 = -10 \, \text{cm} \) (the surface coated with silver, which is treated as a convex surface)

Substituting the values into the lens maker's formula:
\[ \frac{1}{10} = (1.5 - 1) \left( \frac{1}{10} - \frac{1}{-10} \right) \]
\[ \frac{1}{10} = 0.5 \left( \frac{1}{10} + \frac{1}{10} \right) \]
\[ \frac{1}{10} = 0.5 \left( \frac{2}{10} \right) = 0.5 \cdot 0.2 = 0.1 \]

This confirms that the focal length is consistent with the given data.

### Step 3: Use the Lens Formula to Find Image Distance
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( v \) is the image distance
- \( u = -15 \, \text{cm} \)

Rearranging the formula gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{-15} \]
Finding a common denominator (30):
\[ \frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]
Thus,
\[ v = 30 \, \text{cm} \]

### Step 4: Determine the Position of the Image
Since \( v \) is positive, the image is formed on the opposite side of the lens from the object. Therefore, the image is located at \( 30 \, \text{cm} \) from the lens.

### Final Answer
The position of the image is \( 30 \, \text{cm} \) on the opposite side of the lens.

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