Answer :
Let's solve this step-by-step.
### Step 1: Understanding the Initial Conditions
1. Object Distance (Initial): The object is placed at a distance [tex]\( u_i = 10 \, \text{ur} \)[/tex] from the mirror.
2. Initial Image Magnification: The mirror produces a virtual image that is double the size of the object. Therefore, the magnification [tex]\( m_i = 2 \)[/tex].
### Step 2: Finding the Initial Image Distance and Focal Length
For a concave mirror, the magnification [tex]\( m \)[/tex] is given by:
[tex]\[ m = -\frac{v}{u} \][/tex]
Using the initial magnification and object distance:
[tex]\[ 2 = -\frac{v_i}{10} \][/tex]
[tex]\[ v_i = -2 \cdot 10 \][/tex]
[tex]\[ v_i = -20 \, \text{ur} \][/tex]
This indicates that the initial image distance [tex]\( v_i \)[/tex] for the virtual image is [tex]\( -20 \, \text{ur} \)[/tex] (negative value because the image is virtual).
Now, we use the mirror formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Substitute the values for the initial setup:
[tex]\[ \frac{1}{f} = \frac{1}{-20} + \frac{1}{10} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{1}{20} + \frac{1}{10} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{1}{20} + \frac{2}{20} \][/tex]
[tex]\[ \frac{1}{f} = \frac{1}{20} \][/tex]
[tex]\[ f = 20 \, \text{ur} \][/tex]
### Step 3: Finding the Required Conditions for the Real Image
3. Required Image Magnification: The new magnification for the real image is [tex]\( \frac{7}{3} \approx 2.33333 \)[/tex].
From the magnification formula, for the real image:
[tex]\[ m_r = -\frac{v_r}{u_r} \][/tex]
[tex]\[ 2.33333 = -\frac{v_r}{u_i} \][/tex]
[tex]\[ v_r = -2.33333 \cdot u_i \][/tex]
[tex]\[ v_r = -2.33333 \cdot 10 \][/tex]
[tex]\[ v_r \approx -23.3333 \, \text{ur} \][/tex]
### Step 4: Finding the New Object Distance
Using the mirror formula again with the required image distance [tex]\( v_r \)[/tex]:
[tex]\[ \frac{1}{f} = \frac{1}{v_r} + \frac{1}{u_r} \][/tex]
[tex]\[ \frac{1}{20} = \frac{1}{-23.3333} + \frac{1}{u_r} \][/tex]
Solving for [tex]\( u_r \)[/tex]:
[tex]\[ \frac{1}{u_r} = \frac{1}{20} - \frac{1}{-23.3333} \][/tex]
[tex]\[ \frac{1}{u_r} = \frac{1}{20} + \frac{1}{23.3333} \][/tex]
[tex]\[ \frac{1}{u_r} \approx 0.05 + 0.04283 \][/tex]
[tex]\[ \frac{1}{u_r} \approx 0.09283 \][/tex]
[tex]\[ u_r \approx 10.769 \, \text{ur} \][/tex]
### Summary
1. Initial Image Distance: The initial image distance for the virtual image is [tex]\( v_i = -20 \, \text{ur} \)[/tex].
2. Focal Length: The focal length of the mirror is [tex]\( f = 20 \, \text{ur} \)[/tex].
3. Required Image Distance: The required image distance for the real image is [tex]\( v_r \approx -23.3333 \, \text{ur} \)[/tex].
4. Required Object Distance: The object should be placed [tex]\( u_r \approx 10.769 \, \text{ur} \)[/tex].
This means that to achieve a real image that is [tex]\( \frac{7}{3} \approx 2.33333 \)[/tex] times the size of the object, the object should be placed approximately [tex]\( 10.769 \, \text{ur} \)[/tex] from the concave mirror.
### Step 1: Understanding the Initial Conditions
1. Object Distance (Initial): The object is placed at a distance [tex]\( u_i = 10 \, \text{ur} \)[/tex] from the mirror.
2. Initial Image Magnification: The mirror produces a virtual image that is double the size of the object. Therefore, the magnification [tex]\( m_i = 2 \)[/tex].
### Step 2: Finding the Initial Image Distance and Focal Length
For a concave mirror, the magnification [tex]\( m \)[/tex] is given by:
[tex]\[ m = -\frac{v}{u} \][/tex]
Using the initial magnification and object distance:
[tex]\[ 2 = -\frac{v_i}{10} \][/tex]
[tex]\[ v_i = -2 \cdot 10 \][/tex]
[tex]\[ v_i = -20 \, \text{ur} \][/tex]
This indicates that the initial image distance [tex]\( v_i \)[/tex] for the virtual image is [tex]\( -20 \, \text{ur} \)[/tex] (negative value because the image is virtual).
Now, we use the mirror formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Substitute the values for the initial setup:
[tex]\[ \frac{1}{f} = \frac{1}{-20} + \frac{1}{10} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{1}{20} + \frac{1}{10} \][/tex]
[tex]\[ \frac{1}{f} = -\frac{1}{20} + \frac{2}{20} \][/tex]
[tex]\[ \frac{1}{f} = \frac{1}{20} \][/tex]
[tex]\[ f = 20 \, \text{ur} \][/tex]
### Step 3: Finding the Required Conditions for the Real Image
3. Required Image Magnification: The new magnification for the real image is [tex]\( \frac{7}{3} \approx 2.33333 \)[/tex].
From the magnification formula, for the real image:
[tex]\[ m_r = -\frac{v_r}{u_r} \][/tex]
[tex]\[ 2.33333 = -\frac{v_r}{u_i} \][/tex]
[tex]\[ v_r = -2.33333 \cdot u_i \][/tex]
[tex]\[ v_r = -2.33333 \cdot 10 \][/tex]
[tex]\[ v_r \approx -23.3333 \, \text{ur} \][/tex]
### Step 4: Finding the New Object Distance
Using the mirror formula again with the required image distance [tex]\( v_r \)[/tex]:
[tex]\[ \frac{1}{f} = \frac{1}{v_r} + \frac{1}{u_r} \][/tex]
[tex]\[ \frac{1}{20} = \frac{1}{-23.3333} + \frac{1}{u_r} \][/tex]
Solving for [tex]\( u_r \)[/tex]:
[tex]\[ \frac{1}{u_r} = \frac{1}{20} - \frac{1}{-23.3333} \][/tex]
[tex]\[ \frac{1}{u_r} = \frac{1}{20} + \frac{1}{23.3333} \][/tex]
[tex]\[ \frac{1}{u_r} \approx 0.05 + 0.04283 \][/tex]
[tex]\[ \frac{1}{u_r} \approx 0.09283 \][/tex]
[tex]\[ u_r \approx 10.769 \, \text{ur} \][/tex]
### Summary
1. Initial Image Distance: The initial image distance for the virtual image is [tex]\( v_i = -20 \, \text{ur} \)[/tex].
2. Focal Length: The focal length of the mirror is [tex]\( f = 20 \, \text{ur} \)[/tex].
3. Required Image Distance: The required image distance for the real image is [tex]\( v_r \approx -23.3333 \, \text{ur} \)[/tex].
4. Required Object Distance: The object should be placed [tex]\( u_r \approx 10.769 \, \text{ur} \)[/tex].
This means that to achieve a real image that is [tex]\( \frac{7}{3} \approx 2.33333 \)[/tex] times the size of the object, the object should be placed approximately [tex]\( 10.769 \, \text{ur} \)[/tex] from the concave mirror.
