A convergent lens forms a real image 2.41 times the size of the object. The distancebetween object and image is 62.6 cm.Find the distance of the focal point fromthe center of the lens.Answer in units of cm
To find the distance of the focal point from the center of the lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
f = focal length of the lens
v = distance of the image from the center of the lens
u = distance of the object from the center of the lens
We are given the magnification (M), which is the ratio of the height of the image to the height of the object:
M = v/u
In this case, the magnification is given as 2.41, and the distance between the object and image is 62.6 cm.
So, we have:
M = v/u = 2.41
v - u = 62.6 cm
We can rearrange the second equation to get u in terms of v:
u = v - 62.6
Substituting this value of u into the first equation, we have:
M = v/(v - 62.6) = 2.41
Now, let's solve for v.
Multiply both sides of the equation by (v - 62.6):
2.41(v - 62.6) = v
Expand:
2.41v - 151.166 = v
Move all the terms involving v to one side of the equation:
2.41v - v = 151.166
Combine like terms:
1.41v = 151.166
Divide both sides by 1.41:
v = 151.166 / 1.41
v ≈ 107.18 cm
Now, substitute the value of v into the expression for u:
u = v - 62.6
u = 107.18 - 62.6
u ≈ 44.58 cm
Finally, to find the focal length (f), we can use the lens formula with the values of u and v:
1/f = 1/v - 1/u
Substituting the values we found:
1/f = 1/107.18 - 1/44.58
Calculating the values on the right side:
1/f ≈ 0.009317 - 0.022434
1/f ≈ -0.013117
Taking the reciprocal of both sides:
f ≈ -1 / (-0.013117)
f ≈ 76.25 cm
Therefore, the distance of the focal point from the center of the lens is approximately 76.25 cm.