Two lenses of power +2D and -4D. What is the nature of the focal length of each lens An object is placed at a distance of 100 cm from each of the above lenses. Calculate (i) image distance (ii) magnification in each of the two cases
Question
Two lenses of power +2D and -4D. What is the nature of the focal length of each lens An object is placed at a distance of 100 cm from each of the above lenses. Calculate (i) image distance (ii) magnification in each of the two cases
Solution
The power of a lens is the reciprocal of its focal length in meters.
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For the first lens with power +2D, the focal length is 1/2 = 0.5 meters or 50 cm. The positive sign indicates that it is a converging or convex lens.
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For the second lens with power -4D, the focal length is 1/-4 = -0.25 meters or -25 cm. The negative sign indicates that it is a diverging or concave lens.
Now, let's calculate the image distance and magnification for each lens.
(i) Image distance:
We use the lens formula 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.
For the first lens: Rearranging the formula, we get v = 1/(1/f + 1/u) = 1/(1/50 + 1/100) = 1/(0.02 + 0.01) = 33.33 cm. The positive sign indicates that the image is formed on the opposite side of the light source.
For the second lens: v = 1/(1/-25 + 1/100) = 1/(-0.04 + 0.01) = -33.33 cm. The negative sign indicates that the image is formed on the same side as the light source.
(ii) Magnification:
The magnification m is given by the ratio of the image distance to the object distance, m = -v/u.
For the first lens: m = -33.33/-100 = 0.33. The positive sign indicates that the image is upright.
For the second lens: m = 33.33/-100 = -0.33. The negative sign indicates that the image is inverted.
Similar Questions
Find the magnifying power of a compound microscope whose objective and eyepiece are of focal lengths 4.0 cm and 6.0 cm respectively and the object is placed 5.0 cm beyond the objective. Assume that the final image is formed at the least distance of distinct vision (25 cm)
To determine the calculated image distance, we used the formula 𝑑𝑖=𝑑𝑜⋅𝑓1𝑑𝑜−𝑓1d i = d o −f 1 d o ⋅f 1 . For the image distance after the second lens, a similar formula was applied, but instead of using the object distance directly, we used the difference between the distance from the first lens and the location of the image formed by the first lens as the effective object distance. This was expressed as 𝑑𝑜′=𝑑𝑙𝑒𝑛𝑠−𝑑𝑖d o′ =d lens −d i . We also computed the magnification for each lens using both theoretical predictions and experimental data. The theoretical magnification was calculated by dividing the image distance by the object distance. This process was applied to both lenses, with the image formed by the first lens serving as the object for the second lens in our calculations.
An object placed in front of a lens, forms a virtual image whose magnifications is 0.5. The focal length of the lens is 30cm. Calculate the object distance.(with sign)
Power of a concave lens is 5D. An object is placed at a distance of 40cm from the lens. Calculate the image distance(with sign).
The focal length of a lens of power −4D is
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