Question #51837

6 Which of the following is NOT true of experiments involving curved mirrors? a) image distance is negative for for real image b) object distance is positive c) image distance is negative for virtual image d) focal length is negative for convex mirrors
7 An estimate of the refractive index of glass is 1.5. If the angle of incidence is 30o the angle of refraction is? a) 19 o b) 60 o d) 35 o d) 70 o
8 The critical angle for total internal reflection at an air - water interface is approximately 48o. In which of the following situations will total internal reflection occur. a) light incident in water at
40 o b) light incident in water at 55 o c) light incident in air at 40 o d) light incident in air at 55 o
9 When refraction occurs which of the following remains unchanged? a) wave number b) wavelength c) velocity d)frequecy
10 Which of these may NOT be a useful apparatus in an optical experiment? a)optical pins b)lenses c) protractor d) crocodile clip

Expert's answer

Answer on Question 51837, Physics, Optics

6. Which of the following is not true of experiments involving curved mirrors?

a) image distance is negative for real image

b) object distance is positive

c) image distance is negative for virtual image

d) focal length is negative for convex mirrors

Answer:

The image on the convex mirror is always virtual, thus the focal length is negative for convex mirrors. For both types of curved mirrors (convex and concave) the object distance is positive and the image distance is negative for virtual image. The concave mirrors show different images types depending on the distance between the object and the mirror. For example, when object between focus and centre of curvature the image is real, inverted and magnified. But the image distance is positive for real image, not negative.



Thus, the answer is a) image distance is negative for real image

7. An estimate of the refractive index of glass is 1.5. If the angle of incidence is 3030{}^{\circ} the angle of refraction is:

a) 1919{}^{\circ}

b) 6060{}^{\circ}

c) 3535{}^{\circ}

d) 7070{}^{\circ}

Solution:

According to the Snell's law we have:


nisinθi=nrsinθr,n _ {i} \sin \theta_ {i} = n _ {r} \sin \theta_ {r},


where nin_i is the refractive index of air ( ni=1n_i = 1 ), nrn_r is the refractive index of glass ( nr=1.5n_r = 1.5 ), θi\theta_i is the angle of incidence, θr\theta_r is the angle of refraction.

Then, we can find the angle of refraction:


sinθr=ninrsinθi=11.5sin30=0.33,\sin \theta_ {r} = \frac {n _ {i}}{n _ {r}} \sin \theta_ {i} = \frac {1}{1 . 5} \cdot \sin 3 0 {}^ {\circ} = 0. 3 3,θr=arcsin(0.33)=19.2619.\theta_ {r} = \arcsin (0. 3 3) = 1 9. 2 6 {}^ {\circ} \sim 1 9 {}^ {\circ}.


Answer: a) 1919{}^{\circ}

8. The critical angle for total internal reflection at an air-water interface is approximately 4848{}^{\circ} . In which of the following situations will total internal reflection occur?

a) light incident in water at 4040{}^{\circ}

b) light incident in water at 5555{}^{\circ}

c) light incident in air at 4040{}^{\circ}

d) light incident in air at 5555{}^{\circ}

Solution:

The total internal reflection occurs when light attempts to move from a medium having a given refractive index to a medium having a lover refractive index (in our case from water with n1=1.33n_1 = 1.33 to air with n2=1.0n_2 = 1.0 ).



As we can see in the picture for θ1>θc\theta_{1} > \theta_{c} there is no reflected ray. Thus, in order to occur the total internal reflection we need b) light incident in water at 5555{}^{\circ} .

Answer: b) light incident in water at 5555{}^{\circ} .

9. When refraction occurs which of the following remains unchanged?

a) wave number

b) wavelength

c) velocity

d) frequency

Answer:

The law of refraction follows directly from the fact that the speed vv with which light propagates through a medium is inversely proportional to the refractive index of the medium:


v=cn,v = \frac {c}{n},


where, cc is the speed of light in a vacuum, nn is the refractive index of the medium.

Then, from the Snell's law we get:


sinθ1sinθ2=v1v2=n2n1.\frac {\sin \theta_ {1}}{\sin \theta_ {2}} = \frac {v _ {1}}{v _ {2}} = \frac {n _ {2}}{n _ {1}}.


So, while the speed of the light changes when it passes into the new medium the frequency of the light ff remains the same (unchanged). Because, v=cλv = c\lambda for all waves, where λ\lambda is the wavelength, it follows that the wavelength of light must also change as it crosses an interface between two different media. Therefore, the answer is d) frequency.

10. Which of these may not be a useful apparatus in an optical experiment?

a) Optical pins

b) Lenses

c) Protractor

d) Crocodile clip

**Answer:**

We may use optical pins in an optical experiment to illustrate the law of reflection, for example. Also, we can use lenses and protractor. With the help of protractor we can measure the angle of the reflected ray. Thus, the crocodile clip is not useful in an optical experiment and the correct answer is d) Crocodile clip.

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