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Page No 348: - Chapter 9 Ray Optics & Optical Instruments Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


Question 9.22:

At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Answer:

The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m6c222275.jpg

Angle of prism, ∠A = 60°

Refractive index of the prism, µ = 1.524

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m56d2238a.gif  = Incident angle

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_8c6bec3.gif  = Refracted angle

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m5be939b1.gif  = Angle of incidence at the face AC

e = Emergent angle = 90°

According to Snell’s law, for face AC, we can have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m265a03be.gif

It is clear from the figure that anglehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_4190fa5c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m173139fa.gif

According to Snell’s law, we have the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7355/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m109bba20.gif

Hence, the angle of incidence is 29.75°.

Question 9.23:

You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will

(a) deviate a pencil of white light without much dispersion,

(b) disperse (and displace) a pencil of white light without much deviation.

Answer:

(a)Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and white light will emerge from the combination of the two prisms.

(b)Take the system of the two prisms as suggested in Answer (a). Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.

Question 9.24:

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

Answer:

Least distance of distinct vision, d = 25 cm

Far point of a normal eye, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_176b2fd2.gif

Converging power of the cornea, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_65aa0658.gif

Least converging power of the eye-lens, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m3a224d81.gif

To see the objects at infinity, the eye uses its least converging power.

Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 D

Power of the eye-lens is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_72cbeb5f.gif

To focus an object at the near point, object distance (u) = −d = −25 cm

Focal length of the eye-lens = Distance between the cornea and the retina

= Image distance

Hence, image distance, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m18454e0c.gif

According to the lens formula, we can write:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_18939909.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m77b2f56d.gif  = Focal length

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7358/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m3b4d7938.gif

∴Power of the eye-lens = 64 − 40 = 24 D

Hence, the range of accommodation of the eye-lens is from 20 D to 24 D.

Question 9.25:

Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

Answer:

A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye-lens loses its ability of accommodation, the defect is called presbyopia.

Question 9.26:

A myopic person has been using spectacles of power −1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.

Answer:

The power of the spectacles used by the myopic person, P = −1.0 D

Focal length of the spectacles, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7360/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m108a5a8a.gif

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.

During old age, the person uses reading glasses of power,https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7360/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m4b8626c.gif

The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

Question 9.27:

A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Answer:

In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

Question 9.28:

A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Answer:

(a) Focal length of the magnifying glass, f = 5 cm

Least distance of distance vision, d = 25 cm

Closest object distance = u

Image distance, v = −d = −25 cm

According to the lens formula, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7363/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_4b126354.gif

Hence, the closest distance at which the person can read the book is 4.167 cm.

For the object at the farthest distant (u’), the image distance https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7363/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_36db5e5d.gif

According to the lens formula, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7363/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m9aedafc.gif

Hence, the farthest distance at which the person can read the book is

5 cm.

(b) Maximum angular magnification is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7363/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_df025e8.gif

Minimum angular magnification is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7363/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_50b0346c.gif

Question 9.29:

A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(c) Is the magnification in (a) equal to the magnifying power in (b)?

Explain.

Answer:

Note : Here we took focal Length as 10 cm because if we take it as 9 cm then image distance will be zero ,which does not make any sense. (a) Area of each square, A = 1 mm2

Object distance, u = −9 cm

Focal length of a converging lens, =  9 cm

For image distance v, the lens formula can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7365/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_ma6a209e.gif

Magnification, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7365/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_171b29d7.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7365/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m1f119f87.gif

∴Area of each square in the virtual image = (10)2A

= 102 × 1 = 100 mm2

= 1 cm2

(b) Magnifying power of the lens https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7365/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_636efdb6.gif

(c) The magnification in (a) is not the same as the magnifying power in (b).

The magnification magnitude is https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7365/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m15992158.gif and the magnifying power ishttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7365/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m6eadf46a.gif .

The two quantities will be equal when the image is formed at the near point (25 cm).

Question 9.30:

(a) At what distance should the lens be held from the figure in

Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this case?

(c) Is the magnification equal to the magnifying power in this case?

Explain.

Answer:

(a) The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm).

Image distance, v = −d = −25 cm

Focal length, f = 10 cm

Object distance = u

According to the lens formula, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7367/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_3b084e6f.gif

Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.

(b) Magnification =https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7367/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m3af599c6.gif

(c) Magnifying power =https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7367/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m6ad87d71.gif

Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.

Question 9.31:

What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Answer:

Area of the virtual image of each square, A = 6.25 mm2

Area of each square, A0 = 1 mm2

Hence, the linear magnification of the object can be calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7368/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m6685fcb8.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7368/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m17b124c3.gif

Focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/253/7368/NS_3-11-08_Sravana_12_Physics_9_38_NRJ_LVN_html_m5624cc32.gif

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

Also Read : Page-No-349:-Chapter-9-Ray-Optics-&-Optical-Instruments-Exercise-Solutions-class-12-ncert-solutions-Physics

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