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Ray Optics

281921 If a ray travelling in the direction $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ is incident on a plane mirror. After reflection, the ray travels along the direction $\frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$ . The angle of incidence is

1

45^{\circ}

2

75^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

C: Given, Incident direction $= \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$
Reflected direction $= \frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$
Here, the direction of $x$ (i.e. $\hat{i})$ remains same while the direction of $y ($ i.e. $\hat{j})$ change.
Hence, angle of incidence is angle between the unit vector $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ and $\hat{j}$
$\begin{array}{l} ∴ \cos (i) = \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j}) \cdot \hat{j} \\ i = \cos^{- 1} (\frac{\sqrt{3}}{2}) \\ i = 30^{\circ} \end{array}$

CG PET-22.05.2022

Ray Optics

281922 A plane mirror is approaching you at $5 cm . s^{- 1}$ . You can see your image in it. At what speed will your image approach you ?

1

10 cm \cdot s^{- 1}

2

5 cm \cdot s^{- 1}

3

20 cm \cdot s^{- 1}

4

15 cm \cdot s^{- 1}

Explanation:

A: Given, speed of mirror $(V) = 5 cm / \sec$
![original image](https://cdn.mathpix.com/snip/images/FPb-t57BTb-rDt72nlgH2pmCe1Imqw7FAgcB6l9650w.original.fullsize.png)
Let, the distance between the object and the mirror be $d$ As we know, image move 2 times faster than the mirror speed
Image speed $= 2 V = 2 \times 5 = 10 cm / \sec$

AP EAMCET-05.10.2021

Ray Optics

281923 An object is placed at a distance of $40 cm$ in front of a concave mirror of focal length $20 cm$ . The image produced is

1 real, inverted and smaller in size

2 real, inverted and of same size

3 real and erect

4 virtual and inverted

Explanation:

B: Given, focal length (f) $= 20 cm$ and object at a distance of $40 cm$ means object placed at centre of curvature Then,
![original image](https://cdn.mathpix.com/snip/images/4G1VvvK73iAAJqXv06P_ewspB0MxDDTVgpkZPs-r3ro.original.fullsize.png)
Hence, the image produced is real inverted and of same size as that of object.

TS- EAMCET-09.09.2020

Ray Optics

281924 Assertion : Diamond glitters brilliantly.
Reason: Diamond does not absorb sunlight.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-25.05.2019 (M) Shift-1

Ray Optics

281921 If a ray travelling in the direction $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ is incident on a plane mirror. After reflection, the ray travels along the direction $\frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$ . The angle of incidence is

1

45^{\circ}

2

75^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

C: Given, Incident direction $= \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$
Reflected direction $= \frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$
Here, the direction of $x$ (i.e. $\hat{i})$ remains same while the direction of $y ($ i.e. $\hat{j})$ change.
Hence, angle of incidence is angle between the unit vector $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ and $\hat{j}$
$\begin{array}{l} ∴ \cos (i) = \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j}) \cdot \hat{j} \\ i = \cos^{- 1} (\frac{\sqrt{3}}{2}) \\ i = 30^{\circ} \end{array}$

CG PET-22.05.2022

Ray Optics

281922 A plane mirror is approaching you at $5 cm . s^{- 1}$ . You can see your image in it. At what speed will your image approach you ?

1

10 cm \cdot s^{- 1}

2

5 cm \cdot s^{- 1}

3

20 cm \cdot s^{- 1}

4

15 cm \cdot s^{- 1}

Explanation:

A: Given, speed of mirror $(V) = 5 cm / \sec$
![original image](https://cdn.mathpix.com/snip/images/FPb-t57BTb-rDt72nlgH2pmCe1Imqw7FAgcB6l9650w.original.fullsize.png)
Let, the distance between the object and the mirror be $d$ As we know, image move 2 times faster than the mirror speed
Image speed $= 2 V = 2 \times 5 = 10 cm / \sec$

AP EAMCET-05.10.2021

Ray Optics

281923 An object is placed at a distance of $40 cm$ in front of a concave mirror of focal length $20 cm$ . The image produced is

1 real, inverted and smaller in size

2 real, inverted and of same size

3 real and erect

4 virtual and inverted

Explanation:

B: Given, focal length (f) $= 20 cm$ and object at a distance of $40 cm$ means object placed at centre of curvature Then,
![original image](https://cdn.mathpix.com/snip/images/4G1VvvK73iAAJqXv06P_ewspB0MxDDTVgpkZPs-r3ro.original.fullsize.png)
Hence, the image produced is real inverted and of same size as that of object.

TS- EAMCET-09.09.2020

Ray Optics

281924 Assertion : Diamond glitters brilliantly.
Reason: Diamond does not absorb sunlight.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-25.05.2019 (M) Shift-1

Ray Optics

281921 If a ray travelling in the direction $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ is incident on a plane mirror. After reflection, the ray travels along the direction $\frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$ . The angle of incidence is

1

45^{\circ}

2

75^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

C: Given, Incident direction $= \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$
Reflected direction $= \frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$
Here, the direction of $x$ (i.e. $\hat{i})$ remains same while the direction of $y ($ i.e. $\hat{j})$ change.
Hence, angle of incidence is angle between the unit vector $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ and $\hat{j}$
$\begin{array}{l} ∴ \cos (i) = \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j}) \cdot \hat{j} \\ i = \cos^{- 1} (\frac{\sqrt{3}}{2}) \\ i = 30^{\circ} \end{array}$

CG PET-22.05.2022

Ray Optics

281922 A plane mirror is approaching you at $5 cm . s^{- 1}$ . You can see your image in it. At what speed will your image approach you ?

1

10 cm \cdot s^{- 1}

2

5 cm \cdot s^{- 1}

3

20 cm \cdot s^{- 1}

4

15 cm \cdot s^{- 1}

Explanation:

A: Given, speed of mirror $(V) = 5 cm / \sec$
![original image](https://cdn.mathpix.com/snip/images/FPb-t57BTb-rDt72nlgH2pmCe1Imqw7FAgcB6l9650w.original.fullsize.png)
Let, the distance between the object and the mirror be $d$ As we know, image move 2 times faster than the mirror speed
Image speed $= 2 V = 2 \times 5 = 10 cm / \sec$

AP EAMCET-05.10.2021

Ray Optics

281923 An object is placed at a distance of $40 cm$ in front of a concave mirror of focal length $20 cm$ . The image produced is

1 real, inverted and smaller in size

2 real, inverted and of same size

3 real and erect

4 virtual and inverted

Explanation:

B: Given, focal length (f) $= 20 cm$ and object at a distance of $40 cm$ means object placed at centre of curvature Then,
![original image](https://cdn.mathpix.com/snip/images/4G1VvvK73iAAJqXv06P_ewspB0MxDDTVgpkZPs-r3ro.original.fullsize.png)
Hence, the image produced is real inverted and of same size as that of object.

TS- EAMCET-09.09.2020

Ray Optics

281924 Assertion : Diamond glitters brilliantly.
Reason: Diamond does not absorb sunlight.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-25.05.2019 (M) Shift-1

Ray Optics

281921 If a ray travelling in the direction $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ is incident on a plane mirror. After reflection, the ray travels along the direction $\frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$ . The angle of incidence is

1

45^{\circ}

2

75^{\circ}

3

30^{\circ}

4

60^{\circ}

Explanation:

C: Given, Incident direction $= \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$
Reflected direction $= \frac{1}{2} (\hat{i} - \sqrt{3} \hat{j})$
Here, the direction of $x$ (i.e. $\hat{i})$ remains same while the direction of $y ($ i.e. $\hat{j})$ change.
Hence, angle of incidence is angle between the unit vector $\frac{1}{2} (\hat{i} + \sqrt{3} \hat{j})$ and $\hat{j}$
$\begin{array}{l} ∴ \cos (i) = \frac{1}{2} (\hat{i} + \sqrt{3} \hat{j}) \cdot \hat{j} \\ i = \cos^{- 1} (\frac{\sqrt{3}}{2}) \\ i = 30^{\circ} \end{array}$

CG PET-22.05.2022

Ray Optics

281922 A plane mirror is approaching you at $5 cm . s^{- 1}$ . You can see your image in it. At what speed will your image approach you ?

1

10 cm \cdot s^{- 1}

2

5 cm \cdot s^{- 1}

3

20 cm \cdot s^{- 1}

4

15 cm \cdot s^{- 1}

Explanation:

A: Given, speed of mirror $(V) = 5 cm / \sec$
![original image](https://cdn.mathpix.com/snip/images/FPb-t57BTb-rDt72nlgH2pmCe1Imqw7FAgcB6l9650w.original.fullsize.png)
Let, the distance between the object and the mirror be $d$ As we know, image move 2 times faster than the mirror speed
Image speed $= 2 V = 2 \times 5 = 10 cm / \sec$

AP EAMCET-05.10.2021

Ray Optics

281923 An object is placed at a distance of $40 cm$ in front of a concave mirror of focal length $20 cm$ . The image produced is

1 real, inverted and smaller in size

2 real, inverted and of same size

3 real and erect

4 virtual and inverted

Explanation:

B: Given, focal length (f) $= 20 cm$ and object at a distance of $40 cm$ means object placed at centre of curvature Then,
![original image](https://cdn.mathpix.com/snip/images/4G1VvvK73iAAJqXv06P_ewspB0MxDDTVgpkZPs-r3ro.original.fullsize.png)
Hence, the image produced is real inverted and of same size as that of object.

TS- EAMCET-09.09.2020

Ray Optics

281924 Assertion : Diamond glitters brilliantly.
Reason: Diamond does not absorb sunlight.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-25.05.2019 (M) Shift-1

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