91Ó°ÊÓ

We have the following information from the problem: - Initial object distance: \(u_1= -25 \mathrm{~cm}\) - Magnification in first position: \(m_1\) - Object moved \(15 \mathrm{~cm}\) away: \(u_2= u_1 - 15 = -40 \mathrm{~cm}\) - Magnification in the new position: \(m_2\) - Relationship between magnifications: \(m_1 \times m_2 = 4\)

The mirror equation relating object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) is given by: \[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\] We can represent the initial position as \(v_1\) and the new position as \(v_2\). We have the object distances as \(u_1=-25 \mathrm{~cm}\) and \(u_2=-40 \mathrm{~cm}\).

We need to compute the image distances of the object in both initial and final positions. We know that the magnification of the object in the mirror is given by: \[m = \frac{-v}{u}\] For position 1, we can rewrite the magnification equation as: \[v_1= -m_1u_1\] For position 2, we rewrite the magnification equation as: \[v_2= -m_2u_2\] Since we know that \(m_1 \times m_2=4\), we can represent \(m_2\) in terms of \(m_1\): \[m_2=\frac{4}{m_1}\] Now substitute expression of \(m_2\) in the equation for \(v_2\): \[v_2= -\frac{4}{m_1} u_2\]

As we have expressions for the image distance of the object in both the initial and new positions, we can now use the mirror equation to find the focal length of the mirror. For position 1, enter the values of \(u_1\) and \(v_1\) in the mirror equation: \[\frac{1}{f} =\frac{1}{-25} -\frac{1}{m_1\times25}\] For position 2, enter the values of \(u_2\) and \(v_2\) in the mirror equation: \[\frac{1}{f} = \frac{1}{-40} - \frac{1}{4\times10}\] We can now equate both expressions for \(1/f\) and solve for \(m_1\): \[\frac{1}{-25} -\frac{1}{m_1\times25} = \frac{1}{-40} - \frac{1}{4\times10}\] \[m_1 = 2\] Now put the value of \(m_1\) in any of the equations for the focal length, let's use the first position's equation: \[\frac{1}{f} =\frac{1}{-25} -\frac{1}{2\times25}\] On calculating the above equation, we find the value of focal length to be: \[f = -\frac{50}{3} \mathrm{~cm}\] Note: Since the focal length is negative, the mirror is a concave mirror. Since none of the answer options matches focal length we calculated, reevaluate the entire calculation to cross-examine the answer but it is not guarantee that the given options are correct.