Step 1. Understanding the Given Condition: Since the image is real, inverted, and twice the size of the object, we know:
\( m = \frac{v}{u} = -2 \Rightarrow v = -2u \)
Step 2. Set up Equation Using Total Distance: The distance between the object and the image is 45 cm, so:
\( |v - u| = 45 \, \text{cm} \)
Substitute \( v = -2u \) into the equation:
\( |-2u - u| = 45 \)
\( 3|u| = 45 \Rightarrow u = -15 \, \text{cm} \)
Step 3. Determine Image Distance \( v \): Using \( v = -2u \):
\( v = -2 \times (-15) = 30 \, \text{cm} \)
Step 4. Calculate Focal Length Using Lens Formula: Apply the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Substitute \( u = -15 \, \text{cm} \) and \( v = 30 \, \text{cm} \):
\( \frac{1}{f} = \frac{1}{30} - \frac{1}{-15} = \frac{1}{30} + \frac{1}{15} = \frac{1 + 2}{30} = \frac{3}{30} = \frac{1}{10} \)
\( f = +10 \, \text{cm} \)
The problem asks for the focal length of a convex lens that produces a real image, magnified two times, with a distance of 45 cm between the object and the image.
The solution involves the use of the lens formula and the magnification formula for a thin lens. The standard Cartesian sign convention is used, where the optical center is the origin.
1. Lens Formula: The relationship between the object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
2. Magnification Formula: The linear magnification (\(m\)) is given by:
\[ m = \frac{v}{u} \]
For a real image, the magnification is negative (since the image is inverted). For a convex lens, the focal length (\(f\)) is positive, the object distance (\(u\)) is negative, and for a real image, the image distance (\(v\)) is positive.
Step 1: Determine the magnification and relate the image and object distances.
The image is real and magnified two times. A real image formed by a single lens is always inverted. Therefore, the magnification \(m = -2\).
Using the magnification formula:
\[ -2 = \frac{v}{u} \implies v = -2u \]
This relationship is consistent with our sign convention: since \(u\) is negative, \(v\) will be positive.
Step 2: Use the given distance between the object and the image to form a second equation.
The object is placed to the left of the lens (\(u < 0\)), and the real image is formed to the right (\(v > 0\)). The distance between them is the sum of the magnitudes of their positions.
Distance = \(|v| + |u|\). Since \(v > 0\) and \(u < 0\), this is \(v - u\).
\[ v - u = 45 \, \text{cm} \]
Step 3: Solve the two equations to find the object distance (\(u\)) and image distance (\(v\)).
We have a system of two linear equations:
Substitute the first equation into the second:
\[ (-2u) - u = 45 \] \[ -3u = 45 \] \[ u = -15 \, \text{cm} \]
Now, substitute the value of \(u\) back into the first equation to find \(v\):
\[ v = -2(-15) = 30 \, \text{cm} \]
So, the object is 15 cm in front of the lens, and the image is formed 30 cm behind the lens.
Step 4: Use the lens formula to calculate the focal length (\(f\)).
Substitute the values of \(u\) and \(v\) into the lens formula:
\[ \frac{1}{f} = \frac{1}{30} - \frac{1}{-15} \] \[ \frac{1}{f} = \frac{1}{30} + \frac{1}{15} \]
Find a common denominator to add the fractions:
\[ \frac{1}{f} = \frac{1}{30} + \frac{2}{30} \] \[ \frac{1}{f} = \frac{3}{30} = \frac{1}{10} \]
Therefore, the focal length is:
\[ f = 10 \, \text{cm} \]
The focal length of the convex lens is 10 cm.
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.