A convex lens forms an image of an object placed 20cm away from it at a distance of 20cm on the other side of the lens. If the object is moves 5cm toward the lens, the image will be

Options

A. 5cm toward the lens
B. 5cm away from the lens
C. 10cm toward the lens
D. 10cm away from the lens

Explanation

To solve the problem, we will use the lens formula and follow these steps:

### Step 1: Understand the given information
We have a convex lens with the following details:
- The initial object distance (u) = -20 cm (the negative sign indicates that the object is on the same side as the incoming light).
- The image distance (v) = +20 cm (the positive sign indicates that the image is formed on the opposite side of the lens).

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{f} = \frac{1}{20} - \frac{1}{-20} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \]
Thus, the focal length (f) of the lens is:
\[ f = 10 \text{ cm} \]

### Step 3: Update the object distance
The object is moved 5 cm closer to the lens. Therefore, the new object distance (u') is:
\[ u' = -20 \text{ cm} + 5 \text{ cm} = -15 \text{ cm} \]

### Step 4: Calculate the new image distance
Now, we will use the lens formula again to find the new image distance (v'):
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
Substituting the values we have:
\[ \frac{1}{10} = \frac{1}{v'} - \frac{1}{-15} \]
This simplifies to:
\[ \frac{1}{10} = \frac{1}{v'} + \frac{1}{15} \]
To combine the fractions, we need a common denominator, which is 30:
\[ \frac{1}{10} = \frac{3}{30} \quad \text{and} \quad \frac{1}{15} = \frac{2}{30} \]
So we have:
\[ \frac{3}{30} = \frac{1}{v'} + \frac{2}{30} \]
Rearranging gives:
\[ \frac{1}{v'} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]
Thus, the new image distance (v') is:
\[ v' = 30 \text{ cm} \]

### Step 5: Interpret the result
The new image distance of 30 cm indicates that the image has moved further away from the lens compared to the previous position of 20 cm.

### Final Answer
The new image distance is 30 cm, meaning the image has moved 10 cm away from the lens.

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