Question

Consider the situation in figure (18-E7). The bottom of the pot is a reflecting plane mirror, S is a small fish and T is a human eye. Refractive index of water is (a) At what distance(s) from itself will the fish see the image(s) of the eye ? (b) At what distance(s) from itself will the eye see the image(s) of the fish.

Answer 1

Given :

Let x = distance of the image of the eye formed above the surface as seen by the fish

H/x = Real depth / Apparent depth = 1/μ

x= hμ

so distance of direct image = H/2 + μH

= H(μ +1/2)

similarly image through mirror = H/2 +( H+x)

3H/2 +μH

=H(μ+3/2)

b) H/2/ Y= uμ

y= H/2μ

where y= distance of image of fish below the surface  

direct image = H+y

=H+H/2μ

=H(1+1/2μ)Another image of fish will be formed H/2 below mirror

so real depth for hat image becomes =H+H/2= 3H/2

apparent depth = 3H/2μ

so distance of image from eye= H+3H/ 2μ= H(1+3/2μ)

Answer 2

Your answer is explained below:-

Let x = distance of the image of the eye formed above the surface as seen by the fish

H/x = Real depth / Apparent depth = 1/μ

x= hμ

so distance of direct image = H/2 + μH


= H(μ +1/2)


similarly image through mirror = H/2 +( H+x)


3H/2 +μH


=H(μ+3/2)

b) H/2/ Y= uμ

y= H/2μ

where y= distance of image of fish below the surface  

direct image = H+y


=H+H/2μ


=H(1+1/2μ)Another image of fish will be formed H/2 below mirror


so real depth for hat image becomes =H+H/2= 3H/2


apparent depth = 3H/2μ

so distance of image from eye= H+3H/ 2μ= H(1+3/2μ)

Hope it helps @BrainlyHelper