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Two thin lenses of powers 2D and 3D are ...

Two thin lenses of powers `2D` and `3D` are placed in contact. An object is placed at a distance of 30 cm from the combination The distance in cm of the image from the combination is

A

30

B

40

C

50

D

60

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The correct Answer is:
To solve the problem step by step, we will follow the concepts of lens power and the lens formula. ### Step 1: Understanding the Powers of the Lenses We have two thin lenses with powers: - Power of lens 1, \( P_1 = 2D \) - Power of lens 2, \( P_2 = 3D \) ### Step 2: Finding the Total Power of the Combination When two lenses are placed in contact, the total power \( P \) of the combination is the sum of the individual powers: \[ P = P_1 + P_2 = 2D + 3D = 5D \] ### Step 3: Finding the Focal Length of the Combination The focal length \( f \) of a lens is related to its power \( P \) by the formula: \[ f = \frac{100}{P} \quad \text{(in cm)} \] Substituting the total power: \[ f = \frac{100}{5} = 20 \text{ cm} \] ### Step 4: Applying the Lens Formula The lens formula is given by: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Where: - \( v \) is the image distance from the lens, - \( u \) is the object distance from the lens (taken as negative in lens convention). Given that the object distance \( u = -30 \) cm (since the object is placed in front of the lens), we can substitute the values into the lens formula: \[ \frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \] ### Step 5: Rearranging the Lens Formula Rearranging the equation gives: \[ \frac{1}{v} + \frac{1}{30} = \frac{1}{20} \] ### Step 6: Finding a Common Denominator To combine the fractions, we find a common denominator: \[ \frac{1}{v} = \frac{1}{20} - \frac{1}{30} \] The common denominator of 20 and 30 is 60, so we rewrite the fractions: \[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \] Thus: \[ \frac{1}{v} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60} \] ### Step 7: Solving for the Image Distance \( v \) Taking the reciprocal gives: \[ v = 60 \text{ cm} \] ### Conclusion The distance of the image from the combination of the lenses is **60 cm**. ---

To solve the problem step by step, we will follow the concepts of lens power and the lens formula. ### Step 1: Understanding the Powers of the Lenses We have two thin lenses with powers: - Power of lens 1, \( P_1 = 2D \) - Power of lens 2, \( P_2 = 3D \) ### Step 2: Finding the Total Power of the Combination ...
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CP SINGH-REFRACTION AT SPHERICAL SURFACES-EXERCISES
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  2. A convex lens of focal length 40 cm a concave lens of focal length 40 ...

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  3. Two thin lenses of powers 2D and 3D are placed in contact. An object i...

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  4. A real image is formed by a convex lens. If we put a concave lens in c...

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  5. A beam of parallel rays is brought to focus by a planoconvex lens. A t...

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  6. A concave lens and a convex lens have same focal length of 20 cm and b...

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  7. A bi-convex lens is formed with two thin plano-convex lenses as shown ...

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  8. The size of the image of an object, which is at infinity, as formed by...

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  9. Choose the correct option regarding lenses in contact.

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  10. A paralel beam of light incident on a concave lens of focal length 10 ...

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  11. A convex lens A of focal length 20 cm and a concave lens B of focal le...

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  12. Two convex lenses of focal length f1 and f2 are mounted coaxially sepa...

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  13. A combination of two thin lenses with focal lengths f(1) and f(2) resp...

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  14. The position of fical image formed by the given lens combination from ...

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  15. An object is placed 12 cm to the left of a converging lens of focal le...

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  16. A converging lens forms a real image I on its optic axis. A rectangula...

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  17. A lens forms a sharp image on a screen. On inserting a parallel sided ...

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  18. A real image of an object is formed by a conex lens at the bottom of a...

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  19. A concave mirrorr of focal length f(1) is placed at a distance of d fr...

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  20. A concave lens of focal length 20 cm placed in contact with ah plane m...

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