In case of convex lense, when object is moved from f to 2f, its image is real, inverted and magnified. It moves from infinity to 2f on other side.
Focal len is 10 cm. when the object is moved from 15 cm to 25 cm , the magnitude of liner magnification.

Options

A. will increases
B. will decreases
C. will first increases the decreases
D. will first decreases then increases

Answer

To solve the problem step by step, we will use the lens formula and the magnification formula. The given focal length of the convex lens is 10 cm, and we need to find the linear magnification when the object is moved from 15 cm to 25 cm.

### Step 1: Identify the given values
- Focal length (F) = 10 cm
- Initial object distance (u1) = -15 cm (negative because it is on the same side as the incoming light)
- Final object distance (u2) = -25 cm

### Step 2: Use the lens formula to find the image distance (V1) when the object is at u1
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
For the first position (u1 = -15 cm):
\[ \frac{1}{10} = \frac{1}{V1} - \frac{1}{-15} \]
Rearranging gives:
\[ \frac{1}{V1} = \frac{1}{10} + \frac{1}{15} \]
Finding a common denominator (30):
\[ \frac{1}{V1} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} \]
Thus,
\[ V1 = \frac{30}{5} = 6 \text{ cm} \]

### Step 3: Calculate the magnification (M1) for the first position
The magnification is given by:
\[ M = \frac{V}{U} \]
For the first position:
\[ M1 = \frac{V1}{U1} = \frac{6}{-15} = -\frac{2}{5} \]
Taking the magnitude:
\[ |M1| = \frac{2}{5} \]

### Step 4: Use the lens formula to find the image distance (V2) when the object is at u2
For the second position (u2 = -25 cm):
\[ \frac{1}{10} = \frac{1}{V2} - \frac{1}{-25} \]
Rearranging gives:
\[ \frac{1}{V2} = \frac{1}{10} + \frac{1}{25} \]
Finding a common denominator (50):
\[ \frac{1}{V2} = \frac{5}{50} + \frac{2}{50} = \frac{7}{50} \]
Thus,
\[ V2 = \frac{50}{7} \approx 7.14 \text{ cm} \]

### Step 5: Calculate the magnification (M2) for the second position
For the second position:
\[ M2 = \frac{V2}{U2} = \frac{50/7}{-25} = -\frac{2}{7} \]
Taking the magnitude:
\[ |M2| = \frac{2}{7} \]

### Step 6: Compare the magnifications
Now we compare the magnifications:
- Magnification when the object is at 15 cm: |M1| = 2/5 = 0.4
- Magnification when the object is at 25 cm: |M2| = 2/7 ≈ 0.286

Since \(0.4 > 0.286\), we can conclude that the magnification decreases as the object moves from 15 cm to 25 cm.

### Final Answer
The magnitude of linear magnification will **decrease**.

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