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------------------------------------------------ A diverging lens has a focal length of 35.8 cm. If an object is 26 cm from the lens, what is the image distance? Answer in units of cm.

What is the magnitude of the magnification for this lens?

Answer :

The image distance for the object placed 26 cm from the lens is -3.11 cm, and the magnitude of the magnification is 0.12.

The image distance can be calculated using the lens formula:

[tex]1/f = 1/u + 1/v[/tex]

Where f is the focal length of the lens, u is the object distance, and v is the image distance.

Given that the focal length of the diverging lens is 35.8 cm and the object distance is 26 cm, we can substitute these values into the formula:

[tex]1/35.8 = 1/26 + 1/v[/tex]

To solve for v, we need to first find the common denominator:

[tex](26v)/(26v) * (1/35.8) = (26v)/(26v) * (1/26) + (26v)/(26v) * (1/v)[/tex]

Simplifying the equation:

[tex]v/35.8 = 1/26 + 1/v[/tex]

To combine the fractions, we find the common denominator again:

[tex]v/35.8 = (v + 35.8)/26v[/tex]

Cross-multiplying:

[tex]26v^2 = 35.8(v + 35.8)[/tex]

Expanding the equation:

[tex]26v^2 = 35.8v + 35.8^2[/tex]

Rearranging the equation:

[tex]26v^2 - 35.8v - 35.8^2 = 0[/tex]

Using the quadratic formula:

[tex]v = (-b ± √(b^2 - 4ac)) / (2a)[/tex]

[tex]where a = 26, b = -35.8, and c = -35.8^2[/tex]

Plugging in the values:

[tex]v = (-(-35.8) ± √((-35.8)^2 - 4(26)(-35.8^2))) / (2(26))[/tex]

Calculating the equation:


[tex]v = (35.8 ± √(1284.64 + 37729.12)) / 52[/tex]

[tex]v = (35.8 ± √(39013.76)) / 52[/tex]

[tex]v = (35.8 ± 197.52) / 52[/tex]

Therefore, there are two possible solutions:

[tex]v1 = (35.8 + 197.52) / 52 = 233.32 / 52 = 4.48 cm[/tex]
[tex]v2 = (35.8 - 197.52) / 52 = -161.72 / 52 = -3.11 cm[/tex]

Since the object is placed on the same side as the lens, the image distance should be negative. So, the image distance is -3.11 cm.

To find the magnitude of the magnification, we can use the formula:

m = -v/u

Plugging in the values:

[tex]m = -(-3.11) / 26 = 0.12[/tex]

Therefore, the magnitude of the magnification for this lens is 0.12.

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