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Arlecino [84]
2 years ago
7

A convex lens of focal length 25 cm is in contact with a concave lens of focal length 50 cm. The equivalent focal length of this

combination-lenses should be Group of answer choices
Physics
1 answer:
Gwar [14]2 years ago
6 0

The equivalent focal length of this combination-lenses should be +50 cm.

Given:

Focal length of convex lens, f₁ = +25 cm

Focal length of concave lens, f₂ = -50 cm

Calculation:

We know that the power of a lens is given as:

P = 1 / f

where, f is the focal length of lens

Now, the power of convex lens can be calculated as:

P₁ = 1/f₁

  = 1/(25×10⁻² m)

  = 100/25

  = +4 D

Similarly, the power of concave lens can be calculated as:

P₂ = 1/f₂

    = (1/-50×10⁻² m)

    = -100/50

    = -2 D

Thus, the power of the combined lenses can be calculated as:

P₍₁₊₂₎ = P₁ + P₂

       = +4 D - 2 D

       = +2 D

Now, the combined focal length of the lens can be calculated as:

f₍₁₊₂₎ = 1 / P₍₁₊₂₎

      = 100 / 2 D

      = +50 cm

Therefore, the equivalent focal length of this combination-lenses will be +50 cm.

Learn more about mirrors and lenses here:

<u>brainly.com/question/3209252</u>

#SPJ4

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The work done (W) by a constant force (F) is equal to the product of the force in the direction of displacement of the object and the distance (d) moved by the object i.e., W = F * d.

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Positive work - Force acts in the same direction with respect to the displacement of the object. Here, θ is zero, so cos θ i.e., cos 0 is 1. Therefore, from the equation (1), W = Fd (i.e., work done by the force is positive).

Negative work - Force acts in the opposite direction with respect to the displacement of the object.  Here, θ is 180°, so cos θ i.e., cos 180° is -1. Therefore, from the equation (1), W = -Fd (i.e., work done by the force is negative).

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A uniform rod of mass M and length L can pivot freely at one end. Initially, the rod is oriented vertically above the pivot, in
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Answer:

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