Question

9. Locate the image of the point P as seen by the eye in
the figure (18-E5).
H = 1.4
t = 0.4 cm
T
1.0 cm
1
M = 1.3
t = 0.3 cm
T
1.0 cm
1
H = 1.2
t = 0.2 cm
T
1.0 cm
.P
1
Figure 18-E5​

Answer 1

Answer:Solution:

At maximum efficiency, from Fig. 8-5,

CQ = 0.64, CP = 0.60, CH = 0.75

Q = CQnD

3 + 0.64 · 30 s

−1

· (0.5 m)

3 = 2.40 m

3

/s

∆h =

CHn

2D

2

g

=

0.75 · (30 s

−1

)

2

· (0.5 m)2

9.81 m/s

2

= 17.2 m.

P = CPρD

5

n

3 = 0.60 · 1000 kg/m

3

· (0.5 m)5

· (30 s

−1

)

3 = 506 kW.

Explanation:

Answer 2

Answer:

Explanation:

A::B::C

Solution :

The presence of air medium in between the sheet does not affect the shift.

The shift will be due to 3 sheets of different refractive index other than air

=[1−(112)](0.2)+[1−(11.3)](03)

+[1−(11.4)](0.4)

`=0.2 cm above point P.