Focal length of objective of a compound microscope is 4 mm and the image is formed at a distance of 224 mm from it. If angular magnification is 550 , then focal length (in cm) of eyepiece for normal adjustment is

Question

Focal length of objective of a compound microscope is 4 mm and the image is formed at a distance of 224 mm from it. If angular magnification is 550 , then focal length (in cm) of eyepiece for normal adjustment is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

m0=(4-224/4)=-55 m=m0 × me ⇒-550=-55 × me ⇒ me=10 (D/fe)=10 ⇒ fe=(D/10)=2.5 cm

Q. Focal length of objective of a compound microscope is $4 mm$ and the image is formed at a distance of $224 mm$ from it. If angular magnification is $550$ , then focal length (in cm) of eyepiece for normal adjustment is

$m_{0} = \frac{4 - 224}{4} = - 55$
$m = m_{0} \times m_{e}$
$\Rightarrow - 550 = - 55 \times m_{e}$
$\Rightarrow m_{e} = 10$
$\frac{D}{f _{e}} = 10$
$\Rightarrow f_{e} = \frac{D}{10} = 2.5 c m$

Q. Focal length of objective of a compound microscope is 4mm and the image is formed at a distance of 224mm from it. If angular magnification is 550 , then focal length (in cm) of eyepiece for normal adjustment is

m0​=44−224​=−55 m=m0​×me​ ⇒−550=−55×me​ ⇒me​=10 fe​D​=10 ⇒fe​=10D​=2.5cm

Q. Focal length of objective of a compound microscope is $4 mm$ and the image is formed at a distance of $224 mm$ from it. If angular magnification is $550$ , then focal length (in cm) of eyepiece for normal adjustment is

$m_{0} = \frac{4 - 224}{4} = - 55$
$m = m_{0} \times m_{e}$
$\Rightarrow - 550 = - 55 \times m_{e}$
$\Rightarrow m_{e} = 10$
$\frac{D}{f _{e}} = 10$
$\Rightarrow f_{e} = \frac{D}{10} = 2.5 c m$