Answer :
Sure, let's solve this step-by-step.
1. Understand the given values:
- Object distance (u): 30.0 cm
- Image distance (v): 90.0 cm
2. Identify the lens formula:
The lens formula is given by:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
where:
- [tex]\( f \)[/tex] is the focal length,
- [tex]\( v \)[/tex] is the image distance,
- [tex]\( u \)[/tex] is the object distance.
3. Apply the sign convention:
In the lens formula, the object distance (u) is taken as negative for real objects. So, we need to convert the object distance to its negative value:
[tex]\[ u = -30.0 \text{ cm} \][/tex]
The image distance (v) for a real image formed by a converging lens is positive:
[tex]\[ v = 90.0 \text{ cm} \][/tex]
4. Substitute the values into the lens formula:
Let's substitute the values of [tex]\( u \)[/tex] and [tex]\( v \)[/tex] into the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{90.0} - \frac{1}{-30.0} \][/tex]
5. Simplify the equation:
Evaluate the reciprocal values:
[tex]\[ \frac{1}{v} = \frac{1}{90.0} = 0.0111 \text{ (approximately)} \][/tex]
[tex]\[ \frac{1}{u} = \frac{1}{-30.0} = -0.0333 \text{ (approximately)} \][/tex]
Now, substitute these back into the formula:
[tex]\[ \frac{1}{f} = 0.0111 - (-0.0333) \][/tex]
[tex]\[ \frac{1}{f} = 0.0111 + 0.0333 \][/tex]
[tex]\[ \frac{1}{f} = 0.0444 \][/tex]
6. Calculate the focal length (f):
To find [tex]\( f \)[/tex], take the reciprocal of the result:
[tex]\[ f = \frac{1}{0.0444} \][/tex]
Evaluating this gives:
[tex]\[ f \approx 22.5 \text{ cm} \][/tex]
So, the focal length of the converging lens is approximately 22.5 cm.
1. Understand the given values:
- Object distance (u): 30.0 cm
- Image distance (v): 90.0 cm
2. Identify the lens formula:
The lens formula is given by:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
where:
- [tex]\( f \)[/tex] is the focal length,
- [tex]\( v \)[/tex] is the image distance,
- [tex]\( u \)[/tex] is the object distance.
3. Apply the sign convention:
In the lens formula, the object distance (u) is taken as negative for real objects. So, we need to convert the object distance to its negative value:
[tex]\[ u = -30.0 \text{ cm} \][/tex]
The image distance (v) for a real image formed by a converging lens is positive:
[tex]\[ v = 90.0 \text{ cm} \][/tex]
4. Substitute the values into the lens formula:
Let's substitute the values of [tex]\( u \)[/tex] and [tex]\( v \)[/tex] into the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{90.0} - \frac{1}{-30.0} \][/tex]
5. Simplify the equation:
Evaluate the reciprocal values:
[tex]\[ \frac{1}{v} = \frac{1}{90.0} = 0.0111 \text{ (approximately)} \][/tex]
[tex]\[ \frac{1}{u} = \frac{1}{-30.0} = -0.0333 \text{ (approximately)} \][/tex]
Now, substitute these back into the formula:
[tex]\[ \frac{1}{f} = 0.0111 - (-0.0333) \][/tex]
[tex]\[ \frac{1}{f} = 0.0111 + 0.0333 \][/tex]
[tex]\[ \frac{1}{f} = 0.0444 \][/tex]
6. Calculate the focal length (f):
To find [tex]\( f \)[/tex], take the reciprocal of the result:
[tex]\[ f = \frac{1}{0.0444} \][/tex]
Evaluating this gives:
[tex]\[ f \approx 22.5 \text{ cm} \][/tex]
So, the focal length of the converging lens is approximately 22.5 cm.