When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$, the object is to be moved:
When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$, the object is to be moved:
Show Hint
- 40 cm away from the mirror.
- 80 cm away from the mirror.
- 20 cm towards the mirror.
- 20 cm away from the mirror.
The Correct Option is A
Approach Solution - 1
2. Magnification formula: \[ m = \frac{f}{f - u} \] \[ \frac{1}{2} = \frac{f}{f - (-40)} \] \[ f + 40 = 2f \implies f = 40 \mathrm{~cm} \]
3. For magnification $\frac{1}{3}$: \[ \frac{1}{3} = \frac{40}{40 - u} \] \[ 40 - u = 120 \implies u = -80 \mathrm{~cm} \] Therefore, the correct answer is (1) 40 cm away from the mirror.
Approach Solution -2
We are given that when an object is placed 40 cm away from a spherical mirror, an image with magnification \( m_1 = \frac{1}{2} \) is formed. We are asked to find how far the object must be moved to obtain an image with magnification \( m_2 = \frac{1}{3} \).
Concept Used:
For a spherical mirror, the mirror formula is:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
and the magnification \( m \) is given by:
\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \]
Thus, \( v = -mu \).
Step-by-Step Solution:
Step 1: Substitute \( v = -mu \) in the mirror formula.
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{u} + \frac{1}{-mu} \] \[ \frac{1}{f} = \frac{1 - \frac{1}{m}}{u} = \frac{m - 1}{mu} \] \[ f = \frac{mu}{m - 1} \]
Step 2: For the first condition, \( u_1 = 40 \, \mathrm{cm}, \, m_1 = \frac{1}{2} \).
\[ f = \frac{m_1 u_1}{m_1 - 1} = \frac{\frac{1}{2} \times 40}{\frac{1}{2} - 1} \] \[ f = \frac{20}{-\frac{1}{2}} = -40 \, \mathrm{cm} \]
Thus, the mirror has a focal length of \( -40 \, \mathrm{cm} \) (concave mirror).
Step 3: For the new magnification \( m_2 = \frac{1}{3} \), use the same formula to find \( u_2 \).
\[ f = \frac{m_2 u_2}{m_2 - 1} \] Substitute \( f = -40 \): \[ -40 = \frac{\frac{1}{3} u_2}{\frac{1}{3} - 1} = \frac{\frac{1}{3} u_2}{-\frac{2}{3}} = -\frac{u_2}{2} \] \[ u_2 = 80 \, \mathrm{cm} \]
Step 4: Compare the new and old object distances.
\[ u_1 = 40 \, \mathrm{cm}, \quad u_2 = 80 \, \mathrm{cm} \]
The object is moved farther from the mirror by:
\[ \Delta u = u_2 - u_1 = 80 - 40 = 40 \, \mathrm{cm} \]
Final Computation & Result:
Final Answer: The object should be moved 40 cm away from the mirror.
\[ \boxed{\text{Object is moved 40 cm away from the mirror.}} \]
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Optics Questions
- The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be :
- A parallel beam of monochromatic light of wavelength 5000 •Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of ________ (degree).
In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure.
The focal length of the lens is_____cm.
- Two immiscible liquids of refractive indices \( \frac{8}{5} \) and \( \frac{3}{2} \) respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near-normal vision, the apparent depth of the coin is \( \frac{\alpha}{4} \) cm. The value of \( \alpha \) is ______.
- The refractive index of a prism with apex angle \( A \) is \( \cot \frac{A}{2} \). The angle of minimum deviation is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.