An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is:

Using Newton's formula, ⇒ ( f + d 1 ) ( f − d 2 ) = f 2 ⇒ f = d 1 d 2 d 1 − d 2 ⇒ R = 2 d 1 d 2 d 1 − d 2

An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d₁ from C and the distance of the image formed is d₂ from C, the radius of curvature of this mirror is:

Option 1 - <math> <mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>d</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

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Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

Using Newton's formula,

$\Rightarrow (f + d_{1}) (f - d_{2}) = f^{2}$

$\Rightarrow f = \frac{d_{1} d_{2}}{d_{1} - d_{2}} \Rightarrow R = \frac{2 d_{1} d_{2}}{d_{1} - d_{2}}$

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Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. 6.10. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.

(a) While spring is fully compressed all the KE of M₁ is stored as PE of spring.

(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.

(c) If spring is massless, the final state of the M₁ is state of rest.

(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.

a) So spring will continue compress until the blocks moves with the same velocity.

b) As

Payal Gupta

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(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv’

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v’/

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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-

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Time of contact =0.001s

U=126km/h= $\frac{126 \times 1000}{60 \times 60} = \frac{35 m}{s}$

V= -35m/s

Change in momentum of the ball = m (v-u)= $\frac{3}{20} (- 35 - 35) k g m / s$

=21/2

F= dp/dt=- $\frac{21 / 2}{0.001} N$ = - 1.05 $\times 10^{4} N$

Here – negative sign indicates that force will be opposite to the direction of movement of the ball be

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(b) First velocity of the iron sphere increases and after sometimes becomes constant called terminal velocity. Hence according first KE increases and then becomes constant which is best represented by b.

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