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Oksanka [162]
3 years ago
10

4 points

Physics
1 answer:
zubka84 [21]3 years ago
3 0

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

I = k\frac{1}{r^2}

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

I_2 = k\frac{1}{(2r)^2}

Dividing both the equation we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4}  = 25lx

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

I_2 = k\frac{1}{(3r)^2}

Dividing equation 1 and 3 we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9}  = 11.11lx

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

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