Don't tease people having spectacles!

In a concave lens, object (real) distance = 1 m = 1 m and image distance = 2 3 \frac{2}{3} m.

If 4 of these lenses are placed very close to each other ( i.e. having negligible separation between them)

Find the power of the combination. (In dioptres)


The answer is -2.

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1 solution

Aryan Sanghi
Apr 7, 2017

Using the lens formula,

1 f = 1 v 1 u \frac1f = \frac1v - \frac1u

1 f = 1 2 3 1 1 = 1 2 \frac1f = \frac{1}{-\frac23} - \frac{1}{-1}= -\frac{1}{2}

f = 2 m f = -2 m .

Power of 1 lens = 1 f = 0.5 D \text{Power of 1 lens } = \frac{1}{f} = -0.5D

Power of combination = 4 × 0.5 = 2 D \color{#3D99F6}{\boxed{\text{Power of combination } = 4×-0.5 = -2D}}

I have glasses, @Aryan Sanghi !

Yajat Shamji - 11 months, 2 weeks ago

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Even do I!

Aryan Sanghi - 11 months, 2 weeks ago

@Aryan Sanghi , what is the lens' formula? I never read this, please tell me!

Vinayak Srivastava - 11 months, 1 week ago

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According to lens formula, if focal length = f \text{focal length } = f , image distance = v \text{image distance } = v and object distance = u \text{object distance } = u , then

1 f = 1 v 1 u \frac1f = \frac1v - \frac1u

Aryan Sanghi - 11 months, 1 week ago

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OK, thanks a lot! I think you are really good in physics!

Vinayak Srivastava - 11 months, 1 week ago

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@Vinayak Srivastava Thanku for your appreciation. :)

Aryan Sanghi - 11 months, 1 week ago

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