In a compound microscope- the final image is formed at a -160-distance - -d- from the eyepiece- whose focal length is - f-e- -160-where - d gt- D- The object is placed at a distance - -u- from the objective -160-lens- whose focal length -160-is - f-o- The angular magnification -160-of the microscope is- f-ou-f-o- - Dd-1f-e - Df-ou-f-o- - 1d-1f-e - Df-ou-f-o- - 1d-1f-o - Df-ou-f-o- - 1d-1f-e -

A point object O is placed in front of a transparent slab at a distance x from its closer surface. It is seen from the other side fo the slab by light incident nearly normally to the slab. The thickness of the slab is t and its refractive index is $μ .$ Show that the apparent shift in the position fo the object is independent of x and find its value.

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In a compound microscope, the final image is formed at a distance ′d′ from the eyepiece, whose focal length is fe, where d>D. The object is placed at a distance ′u′ from the objective lens, whose focal length is fo. The angular magnification of the microscope is

The correct option is D Dfou+fo×(1d+1fe)angularmagnification=1lateralmagnificationangularmagnification=Dfou+fo×(1d+1fe)option D is correct

In a compound microscope, the final image is formed at a distance $^{'} d^{'}$ from the eyepiece, whose focal length is $f_{e}$ , where $d > D$ . The object is placed at a distance $^{'} u^{'}$ from the objective lens, whose focal length is $f_{o}$ . The angular magnification of the microscope is

The correct option is D $\frac{D f_{o}}{u + f_{o}} \times (\frac{1}{d} + \frac{1}{f_{e}})$
$a n g u l a r_{magnification} = \frac{1}{l a t e r a l_{magnification}}$

$a n g u l a r_{magnification} = \frac{D f_{o}}{u + f_{o}} \times (\frac{1}{d} + \frac{1}{f_{e}})$

option $D$ is correct