An equi-convex lens of focal length F in air is is cut into two halves along the plane perpendicular to the optic axis. One of the halves is placed between a fixed object and a screen separated by distance of 90 cm . Two images are formed on the screen for two different positions of the lens with magnification 2 and 1/2 respectively. Find the value of F . what would be the value of the focal length of the equiconvex lens if the lens is immersed in water ? Refractive index of the material of the lens is 1.5 and that of water is `mu_w=4//3`.

Answer

To solve the problem step by step, we will break down the information provided and apply the relevant formulas.

### Step 1: Understanding the setup
We have an equi-convex lens with focal length \( F \) that is cut into two halves. One half is placed between a fixed object and a screen, which are 90 cm apart. We need to find the focal length \( F \) of the lens based on the given magnifications.

### Step 2: Setting up the magnification equations
Let \( x \) be the distance from the lens to the object. Then the distance from the lens to the image on the screen will be \( 90 - x \).

For the first position where the magnification \( m_1 = 2 \):
\[ m_1 = \frac{v}{u} \implies 2 = \frac{90 - x}{-x} \]
This gives:
\[ 90 - x = -2x \implies 90 = x + 2x \implies 90 = 3x \implies x = 30 \text{ cm} \]

For the second position where the magnification \( m_2 = \frac{1}{2} \):
\[ m_2 = \frac{v}{u} \implies \frac{1}{2} = \frac{90 - x'}{-x'} \]
This gives:
\[ 90 - x' = -\frac{1}{2}x' \implies 90 = x' + \frac{1}{2}x' \implies 90 = \frac{3}{2}x' \implies x' = 60 \text{ cm} \]

### Step 3: Applying the lens formula
Now we can use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the first position:
- \( u = -30 \) cm (object distance is negative)
- \( v = 90 - 30 = 60 \) cm

Substituting these values into the lens formula:
\[ \frac{1}{f} = \frac{1}{60} - \frac{1}{-30} \]
\[ \frac{1}{f} = \frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20} \]
Thus, the focal length \( f = 20 \) cm.

### Step 4: Finding the focal length in water
Using the lens maker's formula:
\[ \frac{1}{f} = (n - 1) \frac{1}{R} \]
Where \( n \) is the refractive index of the lens material. Given \( n = 1.5 \) and \( R \) can be found from the previous calculation:
\[ \frac{1}{20} = (1.5 - 1) \frac{1}{R} \implies \frac{1}{20} = 0.5 \frac{1}{R} \implies R = 10 \text{ cm} \]

Now, when the lens is immersed in water, the effective refractive index becomes:
\[ n_{water} = \frac{n_{lens}}{n_{water}} = \frac{1.5}{\frac{4}{3}} = \frac{1.5 \cdot 3}{4} = \frac{4.5}{4} = 1.125 \]

Applying the lens maker's formula again:
\[ \frac{1}{f_w} = (1.125 - 1) \frac{1}{10} \]
\[ \frac{1}{f_w} = 0.125 \cdot \frac{1}{10} = \frac{0.125}{10} = \frac{1.25}{100} = \frac{1}{80} \]
Thus, the focal length in water \( f_w = 80 \) cm.

### Final Answers
- The focal length \( F \) of the lens in air is **20 cm**.
- The focal length of the lens when immersed in water is **80 cm**.