**QUESTIONS: (3marks each)**
**Q1**. A coin placed at the bottom of a tank appears to be raised when water is poured into it. Explain.

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**Answer 1.** This occurs due to the, phenomenon of refraction of light. Here, the ray of light from the coin travels from a denser medium to a rarer medium . In this process it bends away from the normal . The point from which the refracted rays appear to come gives the apparent position of the coin. As the rays appear to come from a point above the coin, so, the coin seems io be raised.

**Q2**. A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal- length 20 cm. The distance of the object from the len is 30 cm. Find the:

(i) position

(ii) nature

(iii) size of the image formed.

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Answer 2.

Given object size = 5 cm

object distance from lense $u = -30$ cm

focal length $f = 20$ cm,

We have to find $v = ?$

Using the lens formula $\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$

We have,

$ \frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{-30} + \frac{1}{20} = \frac{ -2+3 }{ 60 } = \frac{1}{60}$

or,

$\frac{1}{v} = \frac{1}{60}$

Thus

$v = 60cm$ . This is part (i) of the question

agnification

$\frac{v}{u} = \frac{60}{-30} = -2 $

$Magnification \quad = \quad \frac{Image \quad size}{Object \quad size}$

$\frac{h_i }{h_o}=-2 $

$\frac{h_i}{5}=-2$

$h_{i}= -2 \times 5 = -10 \quad cm$

The image is real inverted and magnified.

**Q3**. (a) Write two rules of the new Cartesian sign conventions for spherical mirrors.

(b) Trace the path of the reflected ray by drawing a figure if it passes from centre of curvature of a concave mirror.

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Answer 3.

(1) New cartesian sign conventions for measuring the various distances in the ray diagrams (reflection by spherical mirrors)

1. All the distances in a ray diagram are measured from the pole of the spherical mirror.

2. The distances measured in the direction of incident light are taken as positive.

3. The distances measured in the direction opposite to the direction of incident light are taken as negative.

4. The heights measured upwards and perpendiculars to the principal axis of the mirror are taken as positive.

5. The heights measured downwards and perpendiculars to the principal axis of the mirror are taken as negative.

Answer 3.

(2) Visit page https://physicscatalyst.com/Class10/reflection_image_formation.php#C1

**Q4**. Design an activity using concave mirror to prove that it s converging in nature. Also state a method to find its rough focal length.

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Answer 4

The following activity can be used to determine the principal focus and approximate focal length of a concave mirror.

(1) Place the mirror vertically on a table.

(2) Place an object (a pencil) in front of the mirror and move it towards and away from the mirror.

(3) Find the approximate spot where the reflection of the object changes from upright to upside down. This point is the focus of the mirror. Measure the distance between this point and the mirror. This length is the focal length of the mirror.

**Q5**. 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of image and magnification. Describe what happens to the image as the needle is moved farther from the mirror.

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Answer 5

Given that $h_1 = 4.5 \quad cm,\quad u = - 12 cm, \quad f = 15 cm$

We have to find $ v = ?$

using the equation of the mirror formula $\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$

we have,

$\frac{1}{15} = \frac{1}{v}-\frac{1}{-12}$

$\frac{1}{15}+\frac{1}{12}=\frac{1}{v}$

By solving we get v= +6.6 cm.

Now, magnification

$m= \frac{h_2}{h_1}=-\frac{v}{u}$

Thus,

$h_2 = - \frac{v}{u} \times h_1 = \frac{-6.6}{-12} \times 4.5 = +2.5 \quad cm$

Hence, magnification of image,

$m=\frac{h_2}{h_1} = \frac{2.5}{4.5} = .56$

The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.

**Q6**. Rohit placed a pencil perpendicular to principal axis in front of a converging mirror of focal length 30 cm. The image formed is twice the size of the pencil. Calculate the distance of the object from the mirror.

**Q7**. Define magnification produced by a spherical mirror in terms of height of a object and image. How is it related to object and image distance? Explain why magnification is positive for virtual image and negative for real image?

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Answer 7

Linear magnification is the ratio of the height of the image to the height of the object. It is represented by the leter $m$

$m=\frac{height \quad of \quad image}{ height \quad of \quad object} = \frac{h_i}{h_o}$

where, $h_i$ is the height of image and $h_o$ is the height of object.

If the image formed is virtual and erect, then the magnification is positive. If the image formed is real and invered, then the magnification is negative.

**Q8**. An object is placed at a distance of 25 cm away from a converging mirror of focal length 20 cm. Discus the effect on the nature and position of the image if the position of the object changes from 25 cm to 15 cm. Justify your answer without using mirror formula.

**Q9**. Write one similarity and one dissimilarity between image formed by pane mirror and convex mirror.

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Answer 9

Similarity :- Both produce Virtual Image .

Disimailarity :- Convex Mirror produces diminished image while plane mirror produce the Image of the same size as that of the object .

**Q10**. (a) State Snell’s law of refraction.

(b) When a ray of light traveling in air enters obliquely into a glass slab, it is observed that the light ray emerges parallel to the incident ray but it is shifted sideways slightly. Draw a ray diagram to illustrate it.

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Answer 10

snell' law of refraction is:- "the ratio of sine of angle of incident to sine of angle of refraction is always constant for given medium"

where,

$\frac{{\sin i}}{{\sin r}} = constan t = {n_{12}}$

where, $n_{12}$ = refractive index of medium 2 with respect to the medium 1.

The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence , all lie in the same plane.

**Q11**. A 5cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.

**Q12**. One half of a convex lens is covered with a black paper.

(a) Show the formation of image of an object placed at 2F1, of such covered lens with the help of ray diagram. Mention the position and nature of image.

(b) Draw the ray diagram for same object at same position in front of the same lens, but now uncovered. Will there be any difference in the image obtained in the two cases? Give reason for your answer.

**Q13**. (a) The refractive index of Ruby is 1.71. What is meant by this statement?

(b) The refractive index of some medium are given below:

Crown glass - 1.52

Water - 1.33

Sapphire - 1.77

In which of the medium is the speed of light

(i) maximum

(ii) minimum
(c) Calculate speed of light in sapphire.

**Q14**. A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30 cm. The distance of the object from the lens is 20 cm. Find the:

(a) Position

(b) Nature

(c) Size of the image formed.

**Q15**. Draw a ray diagram to show refraction of light through a glass slab and label on it the following:

(i) Incident ray

(ii) Refracted ray

(iii) Emergent ray

(iv) Lateral shift (displacement)

**Q16**. If the image formed by mirror for all positions of the object placed in front of it is always virtual and diminished, state the type of the mirror. Draw a ray diagram in support of your answer. Where are such mirrors commonly used and why?

**Q17**. A 6 cm object is placed perpendicular to the principal axis of a convex lens of focal length 15 cm. The distance of the object from the lens is 10 cm. Find the position, size and nature of the mage formed, using the lens formula.

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Class 10 Maths
Class 10 Science