An object of height 3 \(\mathrm{cm}\) is placed at \(25 \mathrm{cm}\) in front of a converging lens of focal length \(20 \mathrm{cm} .\) Behind the lens there is a concave mirror of focal length \(20 \mathrm{cm} .\) The distance between the lens and the mirror is \(5 \mathrm{cm}\). Find the location, orientation and size of the final image.

\( d_i = 100 \mathrm{cm} \) Now let's find the height of the image (\(h_i\)) using magnification formula: \( m = \frac{h_i}{3} = \frac{-100}{25} \) Solving for \(h_i\), we get: \( h_i = -12 \mathrm{cm} \) #Step 2: Calculate the new object distance for the concave mirror# The distance between the lens and the concave mirror is 5 cm. Since the image formed by the lens is at a distance of 100 cm, the light rays need to travel 95 cm back through the lens to reach the mirror. Thus, the new object distance (\(d'_o\)) for the mirror is 95 cm. #Step 3: Calculate the image position and size due to the concave mirror# Now we can use the mirror formula and magnification formula: \( \frac{1}{f} = \frac{1}{d'_o} + \frac{1}{d'_i} \) \( m' = \frac{h'_i}{h_i} = \frac{-d'_i}{d'_o} \) Where \(f\) is the focal length, \(d'_o\) is the new object distance from the mirror, \(d'_i\) is the image distance for the concave mirror, and \(h_i\), \(h'_i\) are the object and image heights involving the mirror respectively. Given, focal length of the mirror (\(f\)) = 20 cm, and new object distance (\(d'_o\)) = 95 cm. We can find the image distance (\(d'_i\)) with the mirror formula: \( \frac{1}{20} = \frac{1}{95} + \frac{1}{d'_i} \) Now, solve for \(d'_i\): \( d'_i = 26.32 \mathrm{cm} \) Now let's find the height of the image (\(h'_i\)) using magnification formula: \( m' = \frac{h'_i}{-12} = \frac{-26.32}{95} \) Solving for \(h'_i\), we get: \( h'_i = 3.32 \mathrm{cm} \) #Result# The final image is located \(26.32 \mathrm{cm}\) in front of the mirror, is upright, and has a height of \(3.32 \mathrm{cm}\).