An object is placed at a distance of 10 cm from a coaxial combination of two lenses A and B contact. The combination forms a real image three times the size of an object. If len B is concave with a focal length of 30 cm, what is the nature and focal lenght of lens A ? a. Convex, 6 cm b. Concave, 12 cm c. Convex, 12 cm d. Convex, 20 cm

Sure, here are the step by step solutions: Given: Linear magnification $m=-3$, u = -10 cm We know that the formula for linear magnification is given by: $m=\frac{v}{u}$ where v is the image distance and u is the object distance. Substituting the given values, we get: $-3=\frac{v}{-10}$ Solving for v, we get: $v=+30cm$ Therefore, the image distance is +30 cm, which means the image is formed 30 cm to the right of the lens. Next, we use the lens formula to find the focal length of the combination of lenses. $\frac{1}{F}=\frac{1}{v}-\frac{1}{u}$ Substituting the values of v and u, we get: $\frac{1}{F}=\frac{1}{30}-\frac{1}{-10}$ Simplifying, we get: $F=\frac{15}{2}cm$ Therefore, the focal length of the combination of lenses is +15/2 cm. Next, we use the lens formula again to find the focal length of the concave lens. $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$ Substituting the value of F and the known focal length of the concave lens (which is negative), we get: $\frac{1}{-\frac{15}{2}}=\frac{1}{-30}+\frac{1}{f_2}$ Simplifying, we get: $f_2=+6cm$ Therefore, the focal length of the concave lens is +6 cm.