If a student desires to produce three times larger and an erect image from a double convex lens with focal length of 15 cm ,then he should place the object from the lens at ;
To determine the position where the object should be placed from the double convex lens in order to produce a three times larger and erect image, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens (positive for an erect image, negative for an inverted image)
u = object distance from the lens
Given:
f = 15 cm
Desired magnification = 3
We know that magnification (m) is given by:
m = v/u
Therefore, we can rewrite the lens formula in terms of magnification:
1/f = 1/v - m/u
Substituting the values:
f = 15 cm
m = 3
1/15 = 1/v - 3/u
Now, we need to find a relationship between v and u to solve the equation. Since the image is three times larger and erect, the magnification (m) will be positive.
m = v/u
3 = v/u
Therefore, we can substitute v/u with 3 in the lens formula:
1/15 = 1/v - 3/u
Now, we can rearrange the equation and solve for u:
1/v = 3/u + 1/15
Multiply both sides by v(3/u + 1/15):
1 = 3v/u + v/15
Combine the terms with v:
1 - v/15 = 3v/u
Multiply both sides by u to eliminate the fraction:
u - v/15u = 3v
Multiply both sides by 15u:
15u^2 - vu = 45v
Rearrange the equation to quadratic form:
15u^2 - vu - 45v = 0
Now we have a quadratic equation. To solve for u, we need to know the value of v or vice versa. Without the specific value of v, we cannot determine the exact position where the object should be placed from the lens to achieve the desired conditions.