Diameter or aperture of a plano - convex lens is 6 cm and its thickness at the center is 3 mm. The image of an object formed is real and twice the size of the object. If the speed of light in the material of the lens is `2xx10^(8)ms^(-1)`. The distance where the object is placed from the plano - convex lens is .............. `xx15cm`.

Answer

To solve the problem step by step, we will follow the information given in the question and the video transcript.

### Step 1: Understand the parameters of the lens
- **Diameter of the lens (D)** = 6 cm
- **Radius of the lens (R)** = D/2 = 3 cm
- **Thickness of the lens at the center (t)** = 3 mm = 0.3 cm

### Step 2: Calculate the radius of curvature (R)
For a plano-convex lens:
\[ R = \frac{r^2}{2t} \]
Where \( r \) is the radius of the lens.
- Convert thickness to cm: \( t = 0.3 \) cm
- Calculate \( R \):
\[ R = \frac{(3 \, \text{cm})^2}{2 \times 0.3 \, \text{cm}} = \frac{9 \, \text{cm}^2}{0.6 \, \text{cm}} = 15 \, \text{cm} \]

### Step 3: Calculate the refractive index (\( \mu \))
Given the speed of light in the material of the lens:
- Speed of light in vacuum \( c = 3 \times 10^8 \, \text{m/s} \)
- Speed of light in the lens material \( v = 2 \times 10^8 \, \text{m/s} \)
\[ \mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 \]

### Step 4: Calculate the focal length (F) of the lens
Using the formula for the focal length of a plano-convex lens:
\[ F = \frac{R}{\mu - 1} \]
Substituting the values:
\[ F = \frac{15 \, \text{cm}}{1.5 - 1} = \frac{15 \, \text{cm}}{0.5} = 30 \, \text{cm} \]

### Step 5: Use the magnification formula
Given that the image is real and twice the size of the object:
\[ \text{Magnification} (m) = \frac{h'}{h} = -\frac{v}{u} = -2 \]
This implies:
\[ v = -2u \]

### Step 6: Apply the lens formula
Using the lens formula:
\[ \frac{1}{F} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( F = 30 \, \text{cm} \) and \( v = -2u \):
\[ \frac{1}{30} = \frac{1}{-2u} - \frac{1}{u} \]
Finding a common denominator:
\[ \frac{1}{30} = -\frac{1}{2u} + \frac{1}{u} = \frac{-1 + 2}{2u} = \frac{1}{2u} \]
Thus:
\[ \frac{1}{30} = \frac{1}{2u} \]
Cross-multiplying gives:
\[ 2u = 30 \implies u = 15 \, \text{cm} \]

### Step 7: Determine the value of dash
From the problem, the distance where the object is placed is given as \( x \times 15 \, \text{cm} \). Since we found \( u = 15 \, \text{cm} \):
\[ x = 1 \]

Thus, the answer is:
\[ \text{Distance where the object is placed from the plano-convex lens} = 15 \, \text{cm} \]

### Final Answer:
The distance where the object is placed from the plano-convex lens is \( 15 \, \text{cm} \).

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