A object is placed before a converging lens. calculate the image formed if the lens as focal length of 15cm also determine the magnification of the length.
Wisdom/Pauline -- please use the same name for your posts.
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again:
https://physicsabout.com/ray-diagrams-for-lenses/
To calculate the image formed by a converging lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the distance of the image from the lens,
u is the distance of the object from the lens.
In this case, the given focal length, f, is 15 cm.
Now, let's assume that the object is placed at a distance u from the lens. We need to convert the value of u into centimeters since the focal length is given in centimeters.
Plugging the values into the lens formula, we have:
1/15 = 1/v - 1/u
Since 1/u is positive, we can write 1/u as 1/u = (v-u)/uv
1/15 = 1/v - (v-u)/uv
To solve for v, we can take the least common denominator:
1/15 = (uv - (v-u))/uv
Multiplying through by 15uv gives:
uv = 15(uv - (v-u))
Distributing and simplifying:
uv = 15uv - 15v + 15u
Now, rearranging the equation:
14uv - 15v = 15u
Next, we need to solve for v. Rearranging the equation gives:
v(14u - 15) = 15u
Dividing both sides by (14u - 15):
v = 15u / (14u - 15)
Since we don't know the specific value of u, we cannot calculate the exact value of v or the size of the image formed without additional information.
However, let's go ahead and determine the magnification, M:
The magnification, M, is given by the formula:
M = -v/u
Plugging in the values:
M = -v/u
M = -(15u / (14u - 15)) / u
Simplifying:
M = -15 / (14u - 15)
So, the magnification of the image formed by the lens with focal length 15 cm is given by the expression -15 / (14u - 15).