A biconvex thick lens is constructed with glass (mu = 1.50). Each of the surfaces has a radius of 10 cm and the thickness at the middle is 5 cm. Locate the image of an object placed far away from the lens.

Explanation

To locate the image formed by a biconvex thick lens, we can follow these steps:

### Step 1: Identify the parameters of the lens
- The lens is biconvex with a refractive index (μ) of 1.50.
- The radius of curvature for both surfaces (R1 and R2) is 10 cm.
- The thickness of the lens at the center is 5 cm.

### Step 2: Determine the object distance for the first surface
Since the object is placed far away from the lens, we can consider the object distance (U) to be infinity (U = ∞).

### Step 3: Apply the lens maker's formula for the first surface
Using the refraction formula at the first surface (Surface 1):
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Where:
- μ2 = 1.5 (glass)
- μ1 = 1 (air)
- R1 = +10 cm (since it is convex)

Substituting the values:
\[ \frac{1.5}{V} - \frac{1}{\infty} = \frac{1.5 - 1}{10} \]
This simplifies to:
\[ \frac{1.5}{V} = \frac{0.5}{10} \]
\[ \frac{1.5}{V} = 0.05 \]
\[ V = \frac{1.5}{0.05} = 30 \text{ cm} \]
So, the image formed by the first surface is at 30 cm from the first surface.

### Step 4: Determine the object distance for the second surface
The image formed by the first surface acts as the object for the second surface. The distance from the first surface to the second surface is the thickness of the lens (5 cm). Therefore, the object distance for the second surface (U2) is:
\[ U_2 = 30 \text{ cm} - 5 \text{ cm} = 25 \text{ cm} \]

### Step 5: Apply the lens maker's formula for the second surface
Using the refraction formula at the second surface (Surface 2):
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Where:
- μ2 = 1 (air)
- μ1 = 1.5 (glass)
- R2 = -10 cm (since it is convex but we take it as negative for the second surface)

Substituting the values:
\[ \frac{1}{V} - \frac{1.5}{25} = \frac{1 - 1.5}{-10} \]
This simplifies to:
\[ \frac{1}{V} - \frac{1.5}{25} = \frac{-0.5}{-10} \]
\[ \frac{1}{V} - \frac{1.5}{25} = 0.05 \]
Now, we can solve for V:
\[ \frac{1}{V} = 0.05 + \frac{1.5}{25} \]
Calculating \(\frac{1.5}{25}\):
\[ \frac{1.5}{25} = 0.06 \]
Thus,
\[ \frac{1}{V} = 0.05 + 0.06 = 0.11 \]
So,
\[ V = \frac{1}{0.11} \approx 9.09 \text{ cm} \]

### Step 6: Conclusion
The final image is formed approximately 9.1 cm from the second surface of the lens.

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