The focal length of a thin lens made from the material of refractive index 1.5 is 15 cm. When it is placed in a liquid of refractive index 4 / 3, its focal length will be cm.

Question

The focal length of a thin lens made from the material of refractive index 1.5 is 15 cm. When it is placed in a liquid of refractive index 4 / 3, its focal length will be cm. (A) 80.31 (B) 50 (C) 78.23 (D) 60

Tardigrade Admin · Accepted Answer

(1/15)=(1.5-1)((1/R1)-(1/R2)) (1/f)=((1.5/4 / 3)-1)((1/R1)-(1/R2)) (f/15)=(1 / 2/(1 / 8)) f =4 × 15=60 cm

Q. The focal length of a thin lens made from the material of refractive index $1.5$ is $15 c m$ . When it is placed in a liquid of refractive index $4/3$ , its focal length will be _______ $c m$ .

80.31

50

78.23

60

$\frac{1}{15} = (1.5 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{f} = (\frac{1.5}{4/3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{f}{15} = \frac{1/2}{( 1/8 )}$
$f = 4 \times 15 = 60 c m$

Q. The focal length of a thin lens made from the material of refractive index 1.5 is 15cm. When it is placed in a liquid of refractive index 4/3, its focal length will be _______cm.

80.31

50

78.23

60

151​=(1.5−1)(R1​1​−R2​1​) f1​=(4/31.5​−1)(R1​1​−R2​1​) 15f​=(1/8)1/2​ f=4×15=60cm

Q. The focal length of a thin lens made from the material of refractive index $1.5$ is $15 c m$ . When it is placed in a liquid of refractive index $4/3$ , its focal length will be _______ $c m$ .

$\frac{1}{15} = (1.5 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{f} = (\frac{1.5}{4/3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{f}{15} = \frac{1/2}{( 1/8 )}$
$f = 4 \times 15 = 60 c m$