An equi-convex lens of focal length 10 cm and refractive index (`mu`g=1.5) is placed in a liquid whose refractive index varies with time as ` mu ( t ) = 1.0 + ( 1 ) /( 10) t `. If the lens was placed in the liquid at t = 0, after what time will the lens act as concave lens of focal length 20 cm?

Options

A. 5 sec
B. 10 sec
C. 15 sec
D. 20 sec

Explanation

To solve the problem step by step, we will use the lens maker's formula and the information provided about the lens and the liquid.

### Step 1: Understand the given parameters
- Focal length of the lens, \( f = 10 \, \text{cm} \)
- Refractive index of the lens, \( \mu_g = 1.5 \)
- Refractive index of the liquid as a function of time, \( \mu(t) = 1 + \frac{1}{10} t \)
- We want to find the time \( t \) when the lens behaves as a concave lens with a focal length of \( -20 \, \text{cm} \).

### Step 2: Use the lens maker's formula
The lens maker's formula for a lens in a medium is given by:
\[ \frac{1}{f} = \mu_g - \mu(t) \cdot \frac{1}{R_1} + \frac{1}{R_2} \]
For an equi-convex lens, \( R_1 = R \) and \( R_2 = -R \). Thus, we can rewrite the formula as:
\[ \frac{1}{f} = (\mu_g - \mu(t)) \cdot \frac{2}{R} \]

### Step 3: Substitute the known values
We know that \( f = 10 \, \text{cm} \) initially, so:
\[ \frac{1}{10} = (1.5 - 1) \cdot \frac{2}{R} \]
This simplifies to:
\[ \frac{1}{10} = 0.5 \cdot \frac{2}{R} \]
From this, we can find \( R \):
\[ \frac{1}{10} = \frac{1}{R} \implies R = 10 \, \text{cm} \]

### Step 4: Set up the equation for when the lens acts as a concave lens
When the lens behaves as a concave lens with a focal length of \( -20 \, \text{cm} \):
\[ \frac{1}{-20} = (1.5 - \mu(t)) \cdot \frac{2}{10} \]
This simplifies to:
\[ -\frac{1}{20} = (1.5 - \mu(t)) \cdot \frac{1}{5} \]

### Step 5: Solve for \( \mu(t) \)
Multiplying both sides by \( -5 \):
\[ \frac{1}{4} = 1.5 - \mu(t) \]
Rearranging gives:
\[ \mu(t) = 1.5 - \frac{1}{4} = 1.25 \]

### Step 6: Set up the equation for \( \mu(t) \)
Now, substituting \( \mu(t) = 1 + \frac{1}{10} t \):
\[ 1 + \frac{1}{10} t = 1.25 \]
Subtracting 1 from both sides:
\[ \frac{1}{10} t = 0.25 \]
Multiplying both sides by 10:
\[ t = 2.5 \, \text{seconds} \]

### Step 7: Conclusion
Thus, the time after which the lens will act as a concave lens of focal length 20 cm is:
\[ \boxed{2.5 \, \text{seconds}} \]