Magnification by a lens of an object at distance 10 cm from it is -2. now a seconnd lens is placed exactly at the same positon where first was kept, without changing the distance between object and lens. The magnification by this second lens is -3.
Now both the lenses are kept in contact at the same place. what will be the new magnification ?

Options

A. `-13/5`
B. `-12/7`
C. `-6/11`
D. `-5/7`

Answer

To solve the problem step by step, we will follow the principles of optics, particularly focusing on magnification and lens formulas.

### Step 1: Understand the Given Information
We have two lenses placed at the same position with an object at a distance of 10 cm from each lens. The magnification for the first lens (M1) is -2, and for the second lens (M2) is -3.

### Step 2: Use the Magnification Formula
The magnification (M) of a lens is given by the formula:
\[ M = \frac{V}{U} \]
where \( V \) is the image distance and \( U \) is the object distance. Since the object distance \( U \) is given as -10 cm (the negative sign indicates that the object is on the same side as the incoming light), we can express the image distances for both lenses.

### Step 3: Calculate Image Distances for Each Lens
For the first lens:
\[ M_1 = -2 = \frac{V_1}{-10} \]
This gives us:
\[ V_1 = -2 \times -10 = 20 \, \text{cm} \]

For the second lens:
\[ M_2 = -3 = \frac{V_2}{-10} \]
This gives us:
\[ V_2 = -3 \times -10 = 30 \, \text{cm} \]

### Step 4: Use the Lens Formula to Find Focal Lengths
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} + \frac{1}{U} \]

For the first lens:
\[ \frac{1}{F_1} = \frac{1}{20} + \frac{1}{-10} \]
\[ \frac{1}{F_1} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = -\frac{1}{20} \]
Thus, \( F_1 = -20 \, \text{cm} \).

For the second lens:
\[ \frac{1}{F_2} = \frac{1}{30} + \frac{1}{-10} \]
\[ \frac{1}{F_2} = \frac{1}{30} - \frac{1}{10} = \frac{1 - 3}{30} = -\frac{2}{30} = -\frac{1}{15} \]
Thus, \( F_2 = -15 \, \text{cm} \).

### Step 5: Combine the Lenses
When two lenses are in contact, the effective focal length \( F_{effective} \) is given by:
\[ \frac{1}{F_{effective}} = \frac{1}{F_1} + \frac{1}{F_2} \]
Substituting the values:
\[ \frac{1}{F_{effective}} = -\frac{1}{20} - \frac{1}{15} \]
To add these fractions, find a common denominator (60):
\[ \frac{1}{F_{effective}} = -\frac{3}{60} - \frac{4}{60} = -\frac{7}{60} \]
Thus, \( F_{effective} = -\frac{60}{7} \, \text{cm} \).

### Step 6: Calculate the New Magnification
Using the lens formula again with the effective focal length:
\[ \frac{1}{F_{effective}} = \frac{1}{V_{new}} + \frac{1}{U} \]
Where \( U = -10 \, \text{cm} \):
\[ -\frac{7}{60} = \frac{1}{V_{new}} - \frac{1}{10} \]
Rearranging gives:
\[ \frac{1}{V_{new}} = -\frac{7}{60} + \frac{1}{10} \]
Converting \( \frac{1}{10} \) to a fraction with a denominator of 60:
\[ \frac{1}{10} = \frac{6}{60} \]
So:
\[ \frac{1}{V_{new}} = -\frac{7}{60} + \frac{6}{60} = -\frac{1}{60} \]
Thus, \( V_{new} = -60 \, \text{cm} \).

Finally, we can find the new magnification \( M_{new} \):
\[ M_{new} = \frac{V_{new}}{U} = \frac{-60}{-10} = 6 \]

### Final Answer
The new magnification when both lenses are kept in contact is \( 6 \).

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