The power of an equi-concave lens is -4.5 D | vedclass

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(D) Power of the equi-concave lens,$P = -4.5 \ D$.Refractive index,$\mu = 1.6$.Focal length of the lens,$f = \frac{1}{P} = \frac{1}{-4.5} \ m = -\frac

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$-26.6$

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(D) Power of the equi-concave lens,$P = -4.5 \ D$.Refractive index,$\mu = 1.6$.Focal length of the lens,$f = \frac{1}{P} = \frac{1}{-4.5} \ m = -\frac{100}{4.5} \ cm = -\frac{200}{9} \ cm$.Using the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For an equi-concave lens,$R_1 = -R$ and $R_2 = R$.Substituting these values into the formula:$\frac{1}{-200/9} = (1.6 - 1) \left( \frac{1}{-R} - \frac{1}{R} ight)$$-\frac{9}{200} = 0.6 \left( -\frac{2}{R} ight)$$\frac{9}{200} = \frac{1.2}{R}$$R = \frac{1.2 	imes 200}{9} = \frac{240}{9} \approx 26.67 \ cm$.Since the question asks for the radius of curvature $R$ of an equi-concave lens,the magnitude is $26.67 \ cm$. Given the sign convention for the concave surface,the radius is typically represented as $-26.6 \ cm$ in this context.

The power of an equi-concave lens is $-4.5 \ D$ and it is made of a material with a refractive index of $1.6$. The radius of curvature of the lens is: (in $cm$)

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