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Lenses. Thin Lens & Magnification Equations. Thin Lens Equation. We can mathematically determine where an image will be formed in relation to the optical centre using the thin lens equation. The Thin Lens Equation. 1 = 1 + 1 f d i d o. d o. Focal length f. F.
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Lenses Thin Lens & Magnification Equations
Thin Lens Equation We can mathematically determine where an image will be formed in relation to the optical centre using the thin lens equation.
The Thin Lens Equation 1 = 1 + 1 f di do do Focal length f F 2F 2F’ F’ di Where: f represents the focal length do represents from the object to the lens di represents from the image to the lens
The Thin Lens Equation – Example 1 • A converging lens has a focal length of 17 cm. A candle is located 48 cm from the lens. What type of image will be formed, and where will it be located?
example 1 (cont'd) G: f = d0 = U: di = ? 17 cm 48 cm
example 1 1 = 1 + 1 f di do E: Rearrange equation: 1 = 1 - 1 di f d0
example 1…almost there! S: 1 = 1 - 1 di 17 cm 48 cm 1 = 0.03799cm-1 di di = 1 / 0.03799cm-1 S: di = 26.327165 cm = 26 cm S: The image of the candle is real and will be about 26 cm from the lens (opposite the object).
example – last step! Sketch the ray diagram for this problem.
Example 2: Thin Lens Equation and Diverging Lenses • A diverging Lens has a focal length of 29 cm. A virtual image of a marble is located 13 cm in front of the lens. Where is the marble (the object) located? • Follow the same steps as before, but be careful with the signs! Refer to your RULES for sign conventions.
Example 2: G: 1 = 1 - 1 d0 -29cm -13cm S: f = - 29 cm di = -13 cm d0 = ? 1 = 0.042444 cm-1 d0 S: U: d0 = 23.560456 cm = 24 cm E: 1 = 1 + 1 f do di S: The marble is located 24 cm from the lens on the same side as the image. 1 = 1 - 1 d0 f di
Example 2: Ray Diagram Sketch • Sketch:
Magnification Equation M = hi = - di ho do ho F 2F 2F’ F’ hi Where: M stands for magnification hi stands for height of the image ho stands for height of the object
Magnification Signs and Measurements • Magnification (M) has no units • M is POSITIVE for an UPRIGHT image • M is NEGATIVE for an INVERTED image If M is…. Then the image is… • GREATER THAN 1 LARGER than the object • BETWEEN 0 AND 1 SMALLER than the object • EQUAL TO 1 SAME SIZE as the object
Example 3: Finding the Magnification of a CONVERGING lens • A toy of height 8.4 cm is balanced in front of a converging lens. An inverted, real image of height 23 cm is noticed on the other side of the lens. What is the magnification of the lens? • Follow the same steps as ex 1 & 2 to solve the problem (just use the magnification equation)
Example 3: G: h0 = 8.4 cm hi = -23 cm M = ? S: M = -23 cm 8.4 cm U: S: M = -2.738095238 = -2.7 E: M = hi ho S: The lens has a magnification of -2.7, which means the image is inverted.
Example 4: Magnification of a Diverging Lens • A coin of height 2.4 cm is placed in front of a diverging lens. An upright, virtual image of height 1.7 cm is noticed on the same side of the lens as the coin. What is the magnification of the lens?
Example 4: G: S: h0 = 2.4 cm hi = 1.7 cm M = ? M = 1.7 cm 2.4 cm U: S: M = 0.708333 = 0.71 E: M = hi ho S: The lens has a magnification of 0.71, which means the image is upright.