An achromatic doublet of focal length 90 cm is to be made of two lens. The material of one having 1.5 times the dispersive power of other The doublet is converging type find focal length of each lens:-

Question

An achromatic doublet of focal length 90 cm is to be made of two lens. The material of one having 1.5 times the dispersive power of other The doublet is converging type find focal length of each lens:- (A) f 1=45 cm f 2=-30 cm (B) f 1=30 cm f 2=-45 cm (C) f 1=-45 cm f 2=60 cm (D) f 1=45 cm f 2=-60 cm

Tardigrade Admin · Accepted Answer

For achromatic doublet (ω1/ f 1)+(ω2/ f 2)=0 ⇒ ( f 1/ f 2)=-(ω l /ω2) (f2/f1)=-1.5 ...(1) (1/ f 1)+(1/ f 2)=(1/90) ...(2) Solving eq.(1) and eq. (2) f 1=30 cm ; f 2=-45 cm

Q. An achromatic doublet of focal length $90 c m$ is to be made of two lens. The material of one having $1.5$ times the dispersive power of other The doublet is converging type find focal length of each lens:-

$f_{1} = 45 c m f_{2} = - 30 c m$

$f_{1} = 30 c m f_{2} = - 45 c m$

$f_{1} = - 45 c m f_{2} = 60 c m$

$f_{1} = 45 c m f_{2} = - 60 c m$

Q. An achromatic doublet of focal length 90cm is to be made of two lens. The material of one having 1.5 times the dispersive power of other The doublet is converging type find focal length of each lens:-

f1​=45cmf2​=−30cm

f1​=30cmf2​=−45cm

f1​=−45cmf2​=60cm

f1​=45cmf2​=−60cm

For achromatic doublet f1​ω1​​+f2​ω2​​=0 ⇒f2​f1​​=−ω2​ωl​​ f1​f2​​=−1.5 ...(1) f1​1​+f2​1​=901​ ...(2) Solving eq.(1) and eq. (2) f1​=30cm;f2​=−45cm

Q. An achromatic doublet of focal length $90 c m$ is to be made of two lens. The material of one having $1.5$ times the dispersive power of other The doublet is converging type find focal length of each lens:-

$f_{1} = 45 c m f_{2} = - 30 c m$

$f_{1} = 30 c m f_{2} = - 45 c m$

$f_{1} = - 45 c m f_{2} = 60 c m$

$f_{1} = 45 c m f_{2} = - 60 c m$