A convex lens forms a real image 6 times the size of the object. The distance between object and image is 33 cm. Find the focal length of the lens.
1/do + 1/di = 1/f
di/do = 6
di + do = 33
You have to solve three equations in three unknowns. Solve the last two in two unknowns first.
do + 6do = 33
do = 33/7 = 4.714 cm,
di = 28.286 cm
Now solve for f using the first equation.
f = 4.0406 cm
To find the focal length of the lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
f = focal length
v = image distance
u = object distance
Given:
v = 33 cm
u = unknown
magnification (image size/object size) = 6
We know that magnification (m) is also defined as -v/u. Rearranging the equation, we have:
m = -v/u
6 = -33/u
Cross-multiplying:
6u = -33
Dividing both sides by 6:
u = -33/6
u = -5.5 cm
Now, let's substitute the values into the lens formula:
1/f = 1/v - 1/u
1/f = 1/33 - 1/(-5.5)
Simplifying the equation:
1/f = 1/33 + 1/5.5
Combining the fractions:
1/f = (1 + 6)/33
1/f = 7/33
Cross-multiplying:
f = 33/7
f = 4.71 cm
Therefore, the focal length of the lens is approximately 4.71 cm.
To find the focal length of the lens, we can use the lens formula:
\(\frac{1}{f} = \frac{1}{d_o} - \frac{1}{d_i}\)
where:
- \(f\) is the focal length of the lens,
- \(d_o\) is the object distance (distance between the object and the lens),
- \(d_i\) is the image distance (distance between the image and the lens).
Given that the image is real and 6 times the size of the object, we have:
\(\frac{h_i}{h_o} = -6\)
where:
- \(h_i\) is the height of the image,
- \(h_o\) is the height of the object.
We are also given that the distance between the object and the image is 33 cm, i.e., \(d_i - d_o = 33\) cm.
To solve for \(f\), we need more information. Let's try to find the magnification (\(M\)) using the equation above for magnification, \(\frac{h_i}{h_o} = -\frac{d_i}{d_o}\).
Given \(h_i = 6h_o\), substituting this into the magnification equation gives:
\(-6 = -\frac{d_i}{d_o}\)
Rearranging the equation, we have:
\(\frac{d_i}{d_o} = 6\)
Since we also know that \(d_i - d_o = 33\), we can solve this system of equations to find \(d_i\) and \(d_o\).
Substituting \(\frac{d_i}{d_o} = 6\) into the equation \(d_i - d_o = 33\), we have:
\(6d_o - d_o = 33\)
\(5d_o = 33\)
\(d_o = \frac{33}{5}\) cm
Substituting this value of \(d_o\) into \(\frac{d_i}{d_o} = 6\), we have:
\(\frac{d_i}{\frac{33}{5}} = 6\)
\(d_i = \frac{33}{5} \times 6\) cm
\(d_i = 39.6\) cm
Now, we can substitute the values of \(d_o\) and \(d_i\) into the lens formula to solve for \(f\):
\(\frac{1}{f} = \frac{1}{\frac{33}{5}} - \frac{1}{39.6}\)
Simplifying the equation, we get:
\(\frac{1}{f} = \frac{5}{33} - \frac{1}{39.6}\)
Finding the common denominator, we have:
\(\frac{1}{f} = \frac{5 \times 39.6 - 1 \times 33}{33 \times 39.6}\)
Calculating the numerator and denominator, we get:
\(\frac{1}{f} = \frac{198 - 33}{1306.8}\)
\(\frac{1}{f} = \frac{165}{1306.8}\)
Simplifying the equation further, we have:
\(\frac{1}{f} = \frac{165}{1306.8}\)
\(f = \frac{1306.8}{165}\)
Calculating this, we find:
\(f \approx 7.92\) cm
Therefore, the focal length of the lens is approximately 7.92 cm.