The text says:
"For the values shown, Vin < +10 volts puts the op-amp output at positive saturation, and VOut= Vin, When Vin exceeds +10 volts the diode closes the feedback loop, clamping the output at 10 volts. "
My inference: When Vin<+10V the opamp increases its output in positive direction in order to equalize voltages at the two inputs..thus as a result the diode become reversed biased and Vin =Vout .. its fine till here... but when Vin>+10V the opamp has to decrease its voltage ..and as the text says that the diode is reversed bias in this case...that means that the opamp outputs some negative voltage (not sure) ..which forwards bias the diode...and the output is clamped to one diode drop below Vopamp.. my problem is ...for Vin>+10V will the opamp output be a negative voltage .... and if yes then suppose Vopamp = -10.7V..for any Vin value greater than 10V.. then -10.7V will experience a voltage drop ....that voltage drop is -0.7V ..
-10.7-(-0.7) = -10V.. ...i.e Vout = -10 V... but it should be 10V...as per text..in order to equalize the inputs....?
//Certainly for Vin>+10V....Vopamp cannot be a positive voltage ..that offcourse will reverse biase the diode again.....which doesnt seems possible..//