Question

biconcave and biconvex lenses with a radius of curvature of 40 cm are kept in contact. the refractive index of the lenses is 4/3 and 3/2 respectively. find the position and nature of the image of the object placed at a distance of 45 cm from the combined lens.

Answer 1

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For BiConvex lens,

R1 = -R2 = 40cm

n = 1.5

F = ?

By using the formula of focal length

1/f = (n-1)[(1/R1) - (1/R2)]

f = 40 cm.

Position of the object = Ui = 20cm

Nature of object = Real and Inverted.

Answer 2

$\huge\blue{Given:}$

R = 40cm

n1 = $\huge{\frac{</strong><strong>4</strong><strong>}{</strong><strong>3</strong><strong>}}$

n2 = $\huge{\frac{</strong><strong>3</strong><strong>}{</strong><strong>2</strong><strong>}}$

u = (-45cm)

$\color {red}\huge\bold\star\underline\mathcal{Question:-}$

We have to find position and nature of image ?

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we know that ,,,

$\frac{n2}{v} - \frac{n1}{u} = \frac{n2 - n1}{r}$

$\frac{(3\div2)}{v} - \frac{(4 \div 3)}{( - 45)} = \frac{(3 \div 2) - (4 \div 3)}{40} \\ \\ \frac{3}{2v} + \frac{4}{135} = \frac{1}{240}$

$\frac{3}{2v} = \frac{ - 55}{2160} \\ \\ \frac{1}{v} \: \: = \frac{ - 11}{584}$

$v \: \: = ( - 53.09)cm$

$so \: the \: image \: is \: behind \: the \: mirror \: and \\ \\ \: the \: image \: is \: virtual \: and \: upright...$

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