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Lenses of power 3D and -5D are combined to form a compound lens. An object is placed at a distance of 50 cm from this lens. The position of its image will be:
A. -10 cm
B. +10 cm
C. -25 cm
D. +25 cm

Last updated date: 13th Jun 2024
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Hint:-In case of thin lenses the power of the combination of lenses are added algebraically & power is equal to 1/focal length of the lens. Find the focal – length and then use the lens formula \[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\] to get the image distance.

Complete step-by-step solution:
Power of a lens is \[P = \dfrac{1}{{focal\;length}}\]
\[ \Rightarrow P = \dfrac{1}{f}m = \dfrac{{100}}{f}cm\]
Also power of combination of two lenses can be added algebraically as \[P = {P_1} + {P_2}\].
\[{P_1} = 3D\]& \[{P_2} = - 5D\]
Power of the combination of two lenses is
\[ \Rightarrow P = 3D - 5D = - 2D\]
Putting the value in the formula
\[ \Rightarrow P = \dfrac{1}{f}m = \dfrac{{100}}{f}cm\]
\[ \Rightarrow P = \dfrac{{100}}{f}cm = - 2D\]
Simplifying for the focal-length , we get
\[ \Rightarrow f = \dfrac{{100}}{{ - 2}} = - 50cm\]
Focal- length of the combination of lenses is -50cm.
Using the lens formula for the combination
\[ \Rightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
where \[v\] is the image distance , \[u\]is the object distance & \[f\]is the focal-length.
Putting the value \[f = - 50cm\;\& \;u = - 50cm\] in the lens formula we get
\[ \Rightarrow \dfrac{1}{v} - \dfrac{1}{{( - 50)}} = \dfrac{1}{{( - 50)}}\]
\[ \Rightarrow \dfrac{1}{v} + \dfrac{1}{{50}} = - \dfrac{1}{{50}}\]
Further simplifying the relation
\[ \Rightarrow \dfrac{1}{v} = - \dfrac{1}{{50}} - \dfrac{1}{{50}} = - \dfrac{2}{{50}}cm\]
\[ \Rightarrow - 2v = 50cm\]
Equating it with the image distance
\[ \Rightarrow v = - 25cm\]
The image is formed at 25 cm in-front of the combination of thin lenses.
Hence option ( C ) is the correct answer.

Note:- A lens has two principal focal lengths which may differ because light can fall on either surface of the lens. The two principal focal lengths are different when medium on two sides have different refractive indices.
If the lower half of a lens is covered with a black paper , the full image of the object is formed because every portion of the lens forms the full image of the object , however the sharpness of the image decreases.