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masya89 [10]
2 years ago
11

A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

s A there is another thin lens, lens B, with a focal length of 27.9 cm . The two lenses share a common axis and are 59.9 cm apart.?
Physics
1 answer:
weeeeeb [17]2 years ago
6 0

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

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A police car drives 100 miles in 75 minutes. What is its average speed in miles per hour?
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A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from
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Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of  incidence  beam striking the water is \theta = 49.63^o

c  the angle of  refraction  beam striking the water is r = 59.7^o

d The angle the refracted beam make with respect to the horizontal is = 30.3^o

e The height of the target above sea level is  

                   h= 125.05m

Explanation:

From the diagram we see that the angle of the beam striking the water is

                   tan \theta = \frac{100}{85}

                        \theta = tan^{-1}(\frac{100}{85} )

                           = 49.63^o

According to Snell's law

                   \mu_{water} *sin(i) = \mu_{air}  *sin(r)

Where \mu_{water } is the refractive index of water =  1.333

           i is the angle of incidence

          \mu_{air} is the refractive index of air  = 1

            r is the angle of refraction

 Substituting values accordingly

          1.33 * sin (40.37) = 1 * sin(r)

    Making r the subject of the formula

                       r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})

                          = 59.7^o

looking at the diagram we can see that to  obtain the angle the refraction beam makes with the horizontal   by subtracting the angle refraction from 90°

                 i.e  90 -59.7 = 30.3 °

From the diagram we see that the height  target above sea level can be obtained by this relation

                   tan \theta = \frac{h}{214}\\

Where h is the is the height

                   tan(30.3) = \frac{h}{214}

                         h = 214 * tan (30.3)

                            =125.05m

           

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