The focal length of a biconvex lens of radii of each surfaces 50 cm and refractive index 1.5, is:

Question

The focal length of a biconvex lens of radii of each surfaces 50 cm and refractive index 1.5, is: (A) 40.4 cm (B) 75 cm (C) 50 cm (D) 80 cm

Tardigrade Admin · Accepted Answer

Using lens makers formula (1/f)=(μ -1)( (1/R1)-(1/R2) ) Here, R1=+ 50 cm, R2=-50cm, μ =1.5 (1/f)=(1.5-1)( (1/50)-(1/(-50)) ) or (1/f)=(1/50) Hence, f=50 cm

Q. The focal length of a biconvex lens of radii of each surfaces 50 cm and refractive index 1.5, is:

40.4 cm

75 cm

50 cm

80 cm

Using lens makers formula $\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ Here, $R_{1} = + 50 c m, R_{2} = - 50 c m, μ = 1.5$ $\frac{1}{f} = (1.5 - 1) (\frac{1}{50} - \frac{1}{( - 50 )})$ or $\frac{1}{f} = \frac{1}{50}$ Hence, $f = 50 c m$

Q. The focal length of a biconvex lens of radii of each surfaces 50 cm and refractive index 1.5, is:

40.4 cm

75 cm

50 cm

80 cm

Using lens makers formula f1​=(μ−1)(R1​1​−R2​1​) Here, R1​=+50cm,R2​=−50cm,μ=1.5 f1​=(1.5−1)(501​−(−50)1​) or f1​=501​ Hence, f=50cm

Using lens makers formula $\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ Here, $R_{1} = + 50 c m, R_{2} = - 50 c m, μ = 1.5$ $\frac{1}{f} = (1.5 - 1) (\frac{1}{50} - \frac{1}{( - 50 )})$ or $\frac{1}{f} = \frac{1}{50}$ Hence, $f = 50 c m$